300+ REAL TIME Class 10 Maths Objective Questions & Answers 2023

[CLASS 10] Mathematics MCQs on Statistics – Mode of Grouped Data

Mathematics Multiple Choice Questions & Answers on “Statistics – Mode of Grouped Data”.

1. What is the mode in a data set?
a) The lowest number in a data set
b) The number that occurs less frequently
c) The number that occurs most frequently
d) The highest number in a data set
Answer: c
Clarification: The number which is occurred most frequently in a dataset is the mode in a data Set. A set of numbers can contain more than one Mode.

2. What is the mode of 8, 5, 7, 10, 15, 21, 5, 7, 2, 5?
a) 2
b) 5
c) 7
d) 21
Answer: b
Clarification: Mode in a data set is defined as the number that occurs most frequently in a data set.
5 occurred three times, 7 occurred two times, 2 and 21 occurred one time each.
5 is the most repeated number according to the definition mode.

3. Below are the results of the students in an exam. Find the mode of given results.
90, 80, 77, 86, 90, 91, 77, 25, 45, 35, 66, 69, 65, 43, 65, 75, 43, 90, 89.
a) 43
b) 77
c) 65
d) 90
Answer: d
Clarification: 90 occurred thrice. 77, 43 and 65 occurred twice.
In the above data set, 90 occurred three times and other numbers appeared once or twice. So, 90 is the mode of the above data set.

4. Find the mode of given samples.
47, 25, 15, 89, 47, 89, 89, 1, 47, 73, 29, 64, 95, 90, 47, 25.
a) 85
b) 47
c) 25
d) No Mode
Answer: b
Clarification: 89 and 25 occurs twice, 64 occurs once, 47 occurs four times.
Since 47 has occurred four times in the above data set and other numbers less than that. The mode of the above data set is 47.

5. What is the mode of 19, 18, 9, 24, 1, 12, 23, 75?
a) 75
b) 1
c) 19
d) No mode
Answer: d
Clarification: According to the definition of mode, a mode is a number that occurs most frequently. There is no such number which is repeated more than once . So, we can say there is No Mode.

6. What is the mode of 20, 15, 13, 12, 9, 17, 13, 1, 9?
a) 13
b) 9
c) Both 13 and 19
d) No mode
Answer: c
Clarification: In the above data set, both 13 and 9 are repeated twice and a data set can have more than one mode. So, the mode is both 13 and 9.

7. Which of the following data sets have the highest Mode?
Data set 1: 11, 13, 14, 25, 30, 14, 14, 4.
Data set 2: 22, 12, 27, 22, 17, 4, 7, 13.
a) Data 1et 2
b) Data set 2
c) Both data sets
d) No mode
Answer: b
Clarification: The mode of the data set 1 is 14 which occurred thrice. The mode of the data set 2 is 22 which occurred twice and the question is which data set has the highest mode so, data set 2 is the answer.

8. The Values of mean and median are given as 22.5 and 20 respectively. Find the approximate value of Mode.
a) 15
b) 20
c) 12
d) 13.5
Answer: a
Clarification: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
Given Mean = 22.5 and Median = 20
22.5 – Mode = 3(22.5 – 20)
Mode = 15

9. Find the value of mode if the mean and median are 25 and 20 respectively.
a) 12
b) 18
c) 10
d) 17.5
Answer: c
Clarification: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
Given Mean = 25 and Median = 20
25 – Mode = 3(25 – 20)
Mode = 10

10. If the value of the mean is 10 and the median is 2 less than the mean. Find the Mode.
a) 2
b) 3
c) 4
d) 5
Answer: c
Clarification: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
Given Mean = 10 and Median = Mean – 2 = 10 – 2 = 8
10 – Mode = 3(10 – 8)
Mode = 4

[CLASS 10] Mathematics MCQs on Solution of Quadratic Equation by Factorisation

Mathematics Multiple Choice Questions and Answers for Class 10 on “Solution of Quadratic Equation by Factorisation”.

1. The sum of a number and its reciprocal is (frac {65}{8}). What is the number?
a) 8
b) 4
c) 2
d) 6
Answer: a
Clarification: Let the number be x
x+(frac {1}{x}=frac {65}{8})
(frac {x^2+1}{x}=frac {65}{8})
8(x²+1)=65x
8x²+8=65x
8x²-65x+8=0
8x²-64x-x+8=0
8x(x-8)-1(x-8)=0
(x-8)(8x-1)=0
x=8, (frac {1}{8})
The number is 8 or (frac {1}{8}).

2. Find two numbers such that the sum of the numbers is 12 and the sum of their squares is 74.
a) 84
b) 75
c) 66
d) 48
Answer: b
Clarification: Sum of the numbers is 12.
Let one number be x. Other number is 12-x.
Sum of their squares = 74
x²+(12-x)²=74
x²+144+x²-24x=74
2x²-24x+144-74=0
2x²-24x+70=0
x²-12x+35=0
x²-7x-5x+35=0
x(x-7)-5(x-7)=0
(x-7)(x-5)=0
x=7, 5
The number is 57 or 75.

3. The sum of two numbers is 13 and the sum of their reciprocals is (frac {13}{40}). What are the two numbers?
a) 76
b) 49
c) 58
d) 94
Answer: c
Clarification: Sum of the numbers is 13.
Let one number be x. Other number is 13-x.
Sum of their reciprocals = (frac {13}{40})
(frac {1}{x} + frac {1}{13-x}=frac {13}{40})
(frac {13-x+x}{x(13-x)}=frac {13}{40})
(frac {13}{13x-x^2}=frac {13}{40})
(frac {1}{13x-x^2}=frac {1}{40})
40=13x-x²
x²-13x+40=0
x²-5x-8x+40=0
x(x-5)-8(x-5)=0
(x-8)(x-5)=0
x=8, 5
The number is 58 or 85.

4. The sum of the squares of the left and right pages of a book is 481. What are the page numbers?
a) 11, 12
b) 12, 13
c) 17, 18
d) 15, 16
Answer: d
Clarification: Since the pages of books are consecutive numbers, so let the left page number be x. The right page number will be x+1.
Sum of the squares of the pages is 481
x²+(x+1)²=481
x²+x²+1+2x=481
2x²+2x-480=0
x²+x-240=0
x²+16x-15x-240=0
x(x+16)-15(x+16)=0
(x-15)(x+16)=0
x=15, -16
Since, page number cannot be negative, so x=15
The two page numbers are 15 and 16.

5. The sum of the squares of two consecutive positive even numbers is 3364. What are the two numbers?
a) 40, 42
b) 38, 40
c) 42, 44
d) 44, 46
Answer: a
Clarification: Let one number be x. The other number is x+2
Sum of the squares of the numbers is 3364.
x²+(x+2)²=3364
x²+x²+4x+4=3364
2x²+4x-3360=0
x²+2x-1680=0
x²+42x-40x-1680=0
x(x+42)-40(x+42)=0
(x+42)(x-40)=0
x=40, -42
Since we only need positive numbers.
Hence, x=40
The two numbers are 40, 42.

6. The sum of the length and the breadth of a rectangle are 97 and the area of the rectangle is 1752. What will be the value of the length and breadth of the rectangle?
a) 42, 76
b) 73, 24
c) 45, 73
d) 22, 77
Answer: b
Clarification: Let the length of the rectangle be x. The sum of length and breadth is 97.
Breadth will be 97-x.
Area of rectangle = 1752.
length × breadth = 1752
x(97-x)=1752
97x-x²=1752
x²-97x+1752=0
x²-73x-24x+1752=0
x(x-73)-24(x-73)=0
(x-73)(x-24)=0
x=73, 24
Hence, the length is 73 and breadth 24 or length 24, breadth 73.

7. The product of digits of a two digit number is 21 and when 36 is subtracted from the number, the digits interchange their places. What is the number?
a) -24
b) 42
c) 73
d) -37
Answer: c
Clarification: Let the units place of the two digit number be x and the tens place be y.
Product of the digits of the places = 21
xy=21
y=(frac {21}{x})
Now, the number will be 10y+x
If 36 is subtracted from the number the digits interchange their places.
New number = 10x+y
10y+x-36=10x+y
9y-9x=36
y-x=4
Now, y=(frac {21}{x})
(frac {21}{x})-x=4
21-x²=4x
x²+4x-21=0
x²+7x-3x-21=0
x(x+7)-3(x+7)=0
(x+7)(x-3)=0
x = -7, 3
The number is 73.

8. If the n^th term of the AP is 5n+2. What will be the value of n so that the sum of the first n terms is 295?
a) 9
b) 10
c) 11
d) 4
Answer: b
Clarification: The n^th term of the AP is 5n+2.
T₁=a=5+2=7
T₂=5(2)+2=12
d=T₂-T₁=12-7=5
S_n=295
S_n=(frac {n}{2})(2a+(n-1)d)
295=(frac {n}{2})(2(7)+(n-1)5)
590=14n+5n²-5n
5n²+9n-590=0
5n²+59n-50n-590=0
n(5n+59)-10(5n+59)=0
(n-10)(5n+59)=0
n=10, (frac {-59}{5})

9. The denominator of a fraction is 1 more than 9 times the numerator. If the sum of the fraction and its reciprocal is (frac {101}{10}) then, what will be the fraction?
a) (frac {89}{10})
b) (frac {10}{89})
c) (frac {1}{10})
d) 10
Answer: c
Clarification: Let the numerator be x.
Denominator of a fraction is 1 more than 9 times the numerator.
Denominator = 1+9x
The fraction is (frac {x}{11+9x})
Fraction + Reciprocal = (frac {101}{10})
(frac {x}{1+9x}+frac {1+9x}{x}=frac {101}{10})
(frac {x^2+(1+9x)^2}{x(1+9x)}=frac {101}{10})
10(x²+1+81x²+18x)=101x(1+9x)
10(82x²+18x+1)=101x+909x²
820x²+180x+10=101x+909x²
89x²-79x-10=0
89x²-89x+10x-10x=0
89x(x-1)+10(x-1)=0
(x-1)(89x+10)=0
x=1, (frac {-10}{89})
The fraction is (frac {1}{10})

10. If the sides of the right angled triangle is x+2, x+1, x then what is the value of x?
a) 1
b) 2
c) 3
d) 4
Answer: a
Clarification: Sides of the right angled triangle is x+2, x+1, x
From Pythagoras theorem,
hyp²=adj²+base²
(x+2)²=(x+1)²+(x)²
x²+4x+4=x²+1+2x+x²
4x+4=x²+1+2x
x²-2x-3=0
x²+3x-x-3=0
x(x+3)-1(x+3)=0
(x-1)(x+3)=0
x=1, -3
Since, sides of triangle cannot be negative.
Hence, x=1

To practice Mathematics Multiple Choice Questions and Answers for Class 10,

[CLASS 10] Mathematics MCQs on Trigonometric Ratios of Complementary Angles

Mathematics Multiple Choice Questions & Answers on “Trigonometric Ratios of Complementary Angles – 1”.

1. Which among these are complementary angles?
a) ∠A + ∠B = 90°
b) ∠A + ∠B = 180°
c) ∠A + ∠B = 60°
d) ∠A + ∠B = 45°
Answer: a
Clarification: Two angles are said to be complementary angles if the sum of these two angles is 90° but if the sum of these two angles is 180° then these two angles are said to be supplementary.

2. The sum of two angles in ∆PQR is complementary with the right angle at Q.
a) True
b) False
Answer: a
Clarification: ∠P + ∠Q + ∠R = 180° ( ∵ The sum of angles in a triangle is 180°)
∠P + 90° + ∠R = 180°
∠P + ∠R = 180° – 90°
∠P + ∠R = 90°
Two angles are said to be complementary angles if the sum of these two angles is 90°.

3. Which trigonometric ratios are positive in the second quadrant?
a) Cosec, Sin
b) Sec, Tan
c) Sin, Cot
d) Tan, Cot
Answer: a
Clarification: A plane is divided into four infinite quadrants. The trigonometric ratios that are positive in the second quadrant are sine, cosecant and the rest of all trigonometric ratios are negative in this quadrant.

4. Sin (90° – x) equals to ______
a) cos x
b) cot x
c) cosec x
d) sec x
Answer: a
Clarification: (90° – x) refers to the first quadrant which lies in the range from 0° to 90°. All trigonometric ratios are positive in the first quadrant and sine changes to cosine when it is 90° or 270°.

5. What is the value of tan 48°?
a) Cot 42°
b) Tan 42°
c) Tan 16°
d) Cot 16°
Answer: a
Clarification: All trigonometric ratios are positive in the first quadrant and tan changes to cot when it is 90° or 270°.
Tan 48° = Tan (90° – 42°)
= Cot 42°

6. Sec 75° equals to _____
a) cosec 15°
b) sec 15°
c) 1
d) 0
Answer: a
Clarification: All trigonometric ratios are positive in the first quadrant and secant changes to cosecant when it is 90° or 270°.
Sec 75° = Sec (90° – 15°)
= Cosec 15°

7. Evaluate (frac {Cot , 54^{circ }}{tan , ⁡36^{circ }}).
a) 0
b) 1
c) (frac {4}{3})
d) (frac {3}{4})
Answer: b
Clarification: (frac {Cot , 54^{circ }}{tan⁡ , 36^{circ }} = frac {Cot , (90^{circ }-36^{circ })}{tan⁡ , 36^{circ }} )
= (frac {tan , 36^{circ }}{tan⁡ , 36^{circ }} )
= 1

8. Find the value of cos 135°.
a) (frac {1}{sqrt {2}})
b) √2
c) -√2
d) (frac {-1}{sqrt {2}})
Answer: d
Clarification: Cos 135° = Cos (90° + 45°)
= -Sin 45°
= (frac {-1}{sqrt {2}})

9. Evaluate tan 75° + cot 65°.
a) Cot 25° + Tan 15°
b) Cot 25° – Tan 15°
c) Cot 15° + Tan 25°
d) Cot 15° – Tan 25°
Answer: c
Clarification: Tan 75° + Cot 65° = Tan (90° – 15°) + Cot (90° – 25°)
= Cot 15° + Tan 25°

10. Cot(180° – a) is _____
a) sine of angle A
b) -cosec of angle A
c) tan of angle A
d) -cot of angle A
Answer: b
Clarification: (180° – a) refers to the second quadrant which lies in the range from 90° to 180°. Trigonometric ratios sine and cosec are only positive in the second quadrant and remaining all the trigonometric ratios are negative.
So, Cot (180° – a) = -Cot A

[CLASS 10] Mathematics MCQs on Statistics – Median of Grouped Data

Mathematics Multiple Choice Questions & Answers on “Statistics – Median of Grouped Data”.

1. What is the median?
a) Difference between higher half and lower half of the data set
b) Mean of the highest and lowest number in a data sample
c) Value separating higher half from the lower half of a data sample
d) Difference between the highest and lowest number.
Answer: c
Clarification: In statistics, the Median is also called the ‘Middle Value’ as it is the value that separates the Highest half from the lower half of a data sample.

2. What is the Median of the following data sample?
2, 7, 4, 8, 9, 10, 6, 12, 13
a) 8
b) 11
c) 9
d) 10
Answer: a
Clarification: Arranging the data sample in ascending order 2, 4, 6, 7, 8, 9, 10, 12, 13.
Median is the middle value which separates the higher half from the lower half in a data sample. After arranging them in ascending order, 8 is the Middle value.

3. What is the Median of the following data sample?
3, 7, 4, 8, 9, 6, 10, 12, 13, 15
a) 7.5
b) 9
c) 8.5
d) 10
Answer: c
Clarification: Arranging the data set in ascending order 3, 4, 6, 7, 8, 9, 10, 12, 13, 15
8 and 9 are the two middle numbers.
Median is the mean of the middle two numbers.
Median is the mean of the middle two numbers = (frac {8+9}{2}) = 8.5

4. Some of the samples are given below. Find the median.
90, 45, 67, 34, 26, 76, 44, 55.
a) 55
b) 45
c) 40
d) 50
Answer: d
Clarification: There is an even number of terms in this Data Set. So, the median will be the mean of the middle two numbers.
Arranging them in ascending order 26, 34, 44, 45, 55, 67, 76, 90.
45 and 55 are the middle two numbers. So, Median = (frac {45+55}{2}) = 50

5. If the mean and the mode are given as 35 and 30. Find the Median.
a) 75
b) 33.33
c) 19
d) 32
Answer: b
Clarification: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
Given Mean = 35, Mode = 30
35 – 30 = 3(35 – Median )
5 = 105 – 3 Median
Median = 33.33

6. If the Mean and Mode are 25, then find the Median.
a) 13
b) 9
c) 25
d) 0
Answer: c
Clarification: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
25 – 25 = 3(25 – Median )
Median = 25

7. What is the formula for the median of Grouped data?
a) Median = L + [(n / 2 – cf) / f] * h
b) Median = L + [(n / 2 + cf) / f] * h
c) Median = L + [(n / 2 – cf) / f] + h
d) Median = L * [(n / 2 – cf) / f] * h
Answer: a
Clarification: Formula for the Median of Grouped Data
Median = L + [(n / 2 – c.f) / f] * h
L = Lower limit of Median Class
cf = Cumulative frequency of the class prior to median class
f = Frequency of Median Class
h = Class size
n = Total frequency

8. Find the Median of the following grouped data.

Marks	Frequency
0-10	9
10-20	10
20-30	24
30-40	16
40-50	11

a) 15
b) 20
c) 26.66
d) 35
Answer: c
Clarification: Total frequency n = 9 + 10 + 24 + 16 + 11 = 70
(frac {n}{2} = frac {70}{2}) = 35

Marks	Frequency	Cumulative Frequency
0-10	9	9
10-20	10	9 + 10 = 19
20-30	24	19 + 24 = 43
30-40	16	43 + 16 = 59
40-50	11	59 + 11 = 70

35 is less than 43 and greater than 19
So, 20 – 30 is the Median Class
Now L = 20, h = 10, cf = 19, f = 24
Median = L + [(n / 2 – c.f) / f] * h
= 20 + [(35 – 19) / 24] * 10
= 26.66

9. Find the Median of given Grouped data.

Rating	Frequency
0-5	12
5-10	20
10-15	10
15-20	6

a) 10
b) 18
c) 8
d) 17.5
Answer: c
Clarification: Total Frequency = 12 + 20 + 10 + 6 = 48 and (frac {n}{2} = frac {48}{2}) = 24

Rating	Frequency	Cumulative Frequency
0-5	12	12
5-10	20	20 + 12 = 32
10-15	10	32 + 10 = 42
15-20	6	42 + 6 = 48

24 is less than 32 and greater than 12. So, the Median Class is 5 – 10.
L = 5, h = 5, cf = 12, f = 20
Median = L + [(n / 2 – c.f) / f] * h
= 5 + [(24 – 12) / 20] * 5
= 8

10. Find the Median of the following grouped data.

Results	Frequency
0-20	5
20-40	10
40-60	30
60-80	15

a) 52
b) 50
c) 34
d) 45
Answer: b
Clarification: Total Frequency n = 5 + 10 + 30 + 15 = 60 and (frac {n}{2} = frac {60}{2}) = 30

Results	Frequency	Cumulative Frequency
0-20	5	5
20-40	10	5 + 10 = 15
40-60	30	15 + 30 = 45
60-80	15	45 + 15 = 60

30 is less than 45 and greater than 15. So, the median class is 40 – 60.
L = 40, h = 20, cf = 15, f = 30
Median = L + [(n / 2 – c.f) / f] * h
= 40 + [(30 – 15) / 30] * 20
= 50

Mathematics MCQs for Class 10 on “Solution of Two Linear Equations in Two Variables in Different Methods”.

1. What will be the nature of the graph lines of the equations 5x-2y+9 and 15x-6y+1?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
Answer: a
Clarification: The given equations are 5x-2y+9 and 15x-6y+1.
Here, a₁=5, b₁=-2, c₁=9 and a₂=15, b₂=-6, c₂=1
Now, (frac {a_1}{a_2} = frac {5}{15} = frac {1}{3}, frac {b_1}{b_2} = frac {-2
}{-6}=frac {1}{3}, frac {c_1}{c_2} = frac {9}{1} )
Clearly, (frac {a_1}{a_2} =frac {b_1}{b_2} ne frac {c_1}{c_2} )
Therefore, the graph lines of the equations will be parallel.

2. What will be the nature of the graph lines of the equations x+3y-2 and 2x-y+5?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
Answer: c
Clarification: The given equations are x+3y-2 and 2x-y+5.
Here, a₁=1, b₁=3, c₁=-2 and a₂=2, b₂=-1, c₂=5
Now, (frac {a_1}{a_2} = frac {1}{2}, frac {b_1}{b_2} = frac {3}{1}) = 3, (frac {c_1}{c_2} = frac {-2}{5} )
Clearly, (frac {a_1}{a_2} ne frac {b_1}{b_2} )
Therefore, the graph lines of the equations will intersect at a point.

3. What will be the nature of the graph lines of the equations 2x+5y+15 and 6x+15y+45?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
Answer: b
Clarification: The given equations are 2x+5y+15 and 6x+15y+45.
Here, a₁=2, b₁=5, c₁=15 and a₂=6, b₂=15, c₂=45
Now, (frac {a_1}{a_2} = frac {2}{6} = frac {1}{3}, frac {b_1}{b_2} = frac {5}{15} =frac {1}{3}, frac {c_1}{c_2} = frac {15}{45} = frac {1}{3} )
Clearly, (frac {a_1}{a_2} =frac {b_1}{b_2} = frac {c_1}{c_2} )
Therefore, the graph lines of the equations will be coincident.

4. What will be the value of k, if the lines given by (5+k)x-3y+15 and (k-1)x-y+19 are parallel?
a) 5
b) 4
c) 6
d) 7
Answer: b
Clarification: The given equations are (5+k)x-3y+15 and (k-1)x-y+19.
Here, a₁=5+k, b₁=-3, c₁=15 and a₂=k-1, b₂=-1, c₂=19
Lines are parallel, so (frac {a_1}{a_2} = frac {b_1}{b_2} ne frac {c_1}{c_2} )
Now, (frac {a_1}{a_2} = frac {5+k}{k-1}, frac {b_1}{b_2} =frac {-3}{-1}) = 3, (frac {c_1}{c_2} = frac {15}{19} )
(frac {5+k}{k-1}) = 3
5+k=3(k-1)
5+k=3k-3
5+3=3k-k
2k=8
k=4

5. What will be the value of k, if the lines given by 3x+ky-4 and 5x+(9+k)y+41 represent two lines intersecting at a point?
a) k≠(frac {7}{2})
b) k≠(frac {27}{8})
c) k=(frac {27}{2})
d) k≠(frac {27}{2})
Answer: d
Clarification: The given equations are 3x+ky-4 and 5x+(9+k)y+41 .
Here, a₁=3, b₁=k, c₁=-4 and a₂=5, b₂=9+k, c₂=41
Lines are intersecting at a point, so (frac {a_1}{a_2} ne frac {b_1}{b_2} )
Now, (frac {a_1}{a_2} =frac {3}{5}, frac {b_1}{b_2} = frac {k}{9+k}, frac {c_1}{c_2} =frac {-4}{41})
(frac {3}{5} ne frac {k}{9+k})
3(9+k)≠5k
27+3k≠5k
27≠5k-3k
2k≠27
k≠(frac {27}{2})

6. What will be the value of k, if the lines given by x+ky+3 and 2x+(k+2)y+6 are coincident?
a) 4
b) 2
c) 6
d) 8
Answer: b
Clarification: The given equations are x+ky+3 and (k-1)x+4y+6.
Here, a₁=1, b₁=k, c₁=3 and a₂=k-1, b₂=4, c₂=6
Lines are coincident, so (frac {a_1}{a_2} =frac {b_1}{b_2} =frac {c_1}{c_2})
Now, (frac {a_1}{a_2} = frac {1}{k-1}, frac {b_1}{b_2} =frac {k}{4}, frac {c_1}{c_2} =frac {3}{6})
(frac {1}{k-1}=frac {k}{4}=frac {1}{2})
2k=4
k=2

7. The lines 5x-7y=13 and 10x-14y=15 are inconsistent.
a) True
b) False
Answer: a
Clarification: A system of linear equations is said to be inconsistent if it has no solution at all.
The given equations are 5x-7y=13 and 10x-14y=15
Here, a₁=5, b₁=-7, c₁=-13 and a₂=10, b₂=-14, c₂=-15
Now, (frac {a_1}{a_2} = frac {5}{10}=frac {1}{2}, frac {b_1}{b_2} =frac {-7}{-14}=frac {1}{2}, frac {c_1}{c_2} =frac {-13}{-15})
Clearly, (frac {a_1}{a_2} =frac {b_1}{b_2} ne frac {c_1}{c_2})
Hence, it will have no solution.
The given equations are inconsistent.

8. The lines 2x+5y=17 and 5x+3y=14 are consistent.
a) False
b) True
Answer: b
Clarification: A system of linear equations is said to be consistent if it has at least one solution.
The given equations are 2x+5y=17 and 5x+3y=14
Here, a₁=2, b₁=5, c₁=-17 and a₂=5, b₂=3, c₂=-14
Now, (frac {a_1}{a_2} =frac {2}{5}, frac {b_1}{b_2} =frac {5}{3}, frac {c_1}{c_2} =frac {-17}{-14})
Clearly, (frac {a_1}{a_2} ne frac {b_1}{b_2} )
Hence, it will have unique solution.
The given equations are consistent.

9. The sum of a two digit number and the number obtained by reversing the order of the digits is 187. If the digits differ by 1, then what will be the number?
a) 67
b) 54
c) 89
d) 67
Answer: c
Clarification: Let the two digit number be 10x+y
The number obtained after reversing the digits will be 10y+x
10x+y+10y+x=187
11x+11y=187
x+y=17 (1)
Also, x-y=1 (2)
Adding (1) and (2)
2x=18
x=9
Substituting in equation (1), 9+y=17
y=8
The number is 89 or 98.

10. It takes 10 men and 6 women to finish a piece of work in 4 days, while it takes 5 men and 7 women to finish the same job in 6 days. What will be the time taken by 1 man and 1 woman to finish the job?
a) Man = 34 days, Woman = 45 days
b) Man = 45 days, Woman = 34 days
c) Man = 53 days, Woman = 96 days
d) Man = 54 days, Woman = 96 days
Answer: d
Clarification: Let 1 man take x days to finish the job and 1 woman take y days to finish the same job.
Then, 1 man’s 1 day work will be (frac {1}{x}) days
1 woman’s 1 day work will be (frac {1}{y}) days
10 men and 6 women can finish the job in 6 days
(10 men’s 1 day work + 6 women’s 1 day work = (frac {1}{4}))
(frac {10}{x}+frac {6}{y}=frac {1}{4})
Let, (frac {1}{x}) = u, (frac {1}{y}) = v
10u+6v=(frac {1}{4}) (1)
5 men and 7 women can finish the job in 6 days.
(5 men’s 1 day work + 7 women’s 1 day work = (frac {1}{6}))
(frac {5}{x}+frac {7}{y}=frac {1}{6})
Let, (frac {1}{x}) = u, (frac {1}{y}) = v
5u + 7v = (frac {1}{6}) (2)
Multiplying equation by 2 and then subtracting both the equations we get,
10u + 14v = (frac {1}{6})
-10u + 6v = (frac {1}{4})
8v = (frac {1}{3}-frac {1}{4})
8v = (frac {1}{12})
v = (frac {1}{96})
v = (frac {1}{y}=frac {1}{96})
y = 96
Substituting the value of v in equation (1) we get,
10u + 6((frac {1}{96})=frac {1}{4})
10u + (frac {1}{16}=frac {1}{4})
10u = (frac {3}{16})
u = (frac {3}{160})
u = (frac {1}{x}=frac {3}{160})
x = (frac {160}{3}) ≈ 54
Hence, a man alone can finish the job in 54 days and a woman can finish the job in 96 days.

11. 10 years ago, a woman was thrice the age of her daughter. Two years later her daughter’s age will be 30 more than the age of the mother. What are the present ages of the woman and the daughter?
a) 70 years, 40 years
b) 60 years, 40 years
c) 55 years, 25 years
d) 45 years, 20 years
Answer: c
Clarification: Let the present ages of mother be x years and daughter be y years.
10 years ago,
Age of mother = x-10 years
Age of daughter = y-10 years
Age of mother = 3(age of daughter)
x-10=3(y-10)
x-10=3y-30
x-3y+20=0
x=3y-20 (1)
Two years later,
Age of mother = x+2 years
Age of daughter = y+2 years
Age of mother will be 30 more than the age of daughter
x+2=y+2+30
x=y+30 (2)
From (1) and (2), we get,
y+30=3y-20
30+20=3y-y
50=2y
y=25
Substituting y=25 in equation (1) we get,
x=3(25)-20
x=55
The present age of mother is 55 years and that of daughter is 25 years.

12. The sum of two numbers is 13 and the sum of their reciprocals is (frac {13}{40}). What are the two numbers?
a) 5, 8
b) 10, 3
c) 12, 1
d) 9, 4
Answer: a
Clarification: Let the two numbers be x and y.
x+y=13 (1)
Also, (frac {1}{x}+frac {1}{y}=frac {13}{40})
(frac {y+x}{xy}=frac {13}{40})
40(y+x)=13xy
xy=40
Now, x-y=(sqrt {(x+y)^2-4xy})
=(sqrt {13^2-4(40)})
=(sqrt {169-160})
=√9
=±3
x-y=3 or x-y=-3 (2)
Adding (1) and (2), we get,
x+y=13
-x-y=3
2y=16
y=8
x=5
Or
y=5
x=8

13. In a piggy bank the total number of coins of Rs. 5 and Rs. 1 is 100. If the total coins amount is 300, then what is the number of coins of each denomination?
a) 30, 70
b) 50, 50
c) 45, 55
d) 60, 40
Answer: b
Clarification: Let the coins of Rs. 5 be x and that of Rs. 1 be y
Total number of coins is 100
x+y=100
y=100-x
Total amount of coins is 300
Also, 5x+y=300 (1)
Substituting y=100-x in equation (1) we get,
5x+100-x=300
4x=200
x=50
y=100-x=100-50=50
The coins of each denomination are 50.

14. A father gives Rs. 500 to his children every month. If the boy gets Rs. 100 then, the girl gets Rs. 200 and if the boy gets Rs. 100 the girl gets Rs. 150. How many children does he have?
a) 0
b) 3
c) 2
d) 1
Answer: b
Clarification: Let the number of boys be x and number of girls be y
Now, if the boys get 100, the girls get 200
100x+200y=500
x+2y=5
x=5-2y
If the boys get 200, the girls get 150
200x+150y=500
4x+3y=10 (1)
Substituting x=5-2y in equation 1 we get,
4(5-2y)+3y=10
20-8y+3y=10
-5y=-10
y=2
x=5-2y=5-2(2)=1
The father has three children.

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