FREE RRB NTPC MATHEMATICS MOCK TEST

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FREE RRB NTPC MATHEMATICS MOCK TEST

Preparing for the RRB (Railway) exam? Maths is an important part, and the best way to get better is by practicing. Try our free RRB Mathematics mock test to check how much you know, find out where you need to improve, and build your confidence. The test includes all the main topics like number system, ratio and proportions, percentages, time & work, and more. It’s just like the real exam. After the test, you’ll get answers, explanations, and tips to help you do better.

LCM and HCF quiz

LCM AND HCF QUIZ

Total number of questions 10

1 / 10

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

2 / 10

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes)

In 30 minutes they will toll together [ 30/2] +1 =16 times

3 / 10

What is the largest three-digit number that is exactly divisible by the HCF of 24 and 36?

990

996

100

900

The HCF of 24 and 36 is 12. To find the largest three-digit number that is exactly divisible by 12, we need to find the largest multiple of 12 that is less than 1000.

The largest multiple of 12 less than 1000 is 996 (83 × 12 = 996).

4 / 10

Find the largest number that divides 64, 136, and 238 to leave the same remainder in each case.

To find the required number, we need to find the HCF of (136-64), (238-136), and (238-64), i.e., HCF (72, 102, 174).

72 = 2³ x 3²
102 = 2 x 3 x 17
174 = 2 x 3 x 29

Therefore, HCF (72, 102, 174) = 2 x 3 = 6

Hence, 6 is the required number.

5 / 10

William and Nancy are friends as well as football coaches in the same school. William goes to
Martyr Day school every 3 days while Nancy goes there every 5 days to deliver coaching classes.
If they both delivered the coaching sessions today, how many days, in the next 60 days, will both of
them be delivering the coaching sessions on the same day?

In order to find the number of days after which both of them will go together, we need to find a
number that is as small as possible and is divisible by both 3 and 5.
In other words, we need to calculate the LCM of 3 and 5 which is: 15 days.
Now, we know that both of them go to school together every 15th
day.
Hence, the number of days, in the next 60 days, when both of them will deliver sessions on
the same day =60/15
= 4 days

6 / 10

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

7 / 10

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

8 / 10

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

120

200

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

9 / 10

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

26 minutes 18 seconds

42 minutes 36 seconds

45 minutes

46 minutes and 12 seconds

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

10 / 10

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

504

536

544

548

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

Your score is

The average score is 78%

NUMBER SYSTEM SIMPLIFICATION QUIZ

Created on December 08, 2021 By

Bina singh

PEMDAS QUIZ

1 / 10

14+18 ÷2 x 18 -7

169

2 / 10

15 x 18 +12 ÷ 3 +9

283

564

429

3 / 10

8 x 4 +9 -9+18

4 / 10

11 x 11 - 6 x 17 +4

5 / 10

$(6+3)^{2}$ +(9-10 ÷ 5)

306

6 / 10

$(10+59-3^{2})$ ÷ (24-4)

100

7 / 10

2-20÷ 5 x 3

-10

-51

8 / 10

$4 \times (12\times 6-4^{2}) +9$

467

233

569

9 / 10

(19-8) x (10+4) + $8^{2}$

218

439

10 / 10

(10+43-5) ÷6 + $5^{2}$

Your score is

The average score is 72%

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NUMBER SYSTEM BODMAS

213

Created on December 11, 2021 By

Bina singh

BODMAS QUIZ CLASS 5

1 / 10

(10+43-5) ÷6 + $5^{2}$

2 / 10

(19-8) x (10+4) + $8^{2}$

218

439

3 / 10

2-20÷ 5 x 3

-10

-51

4 / 10

$(10+59-3^{2})$ ÷ (24-4)

100

5 / 10

14+18 ÷2 x 18 -7

169

6 / 10

8 x 4 +9 -9+18

7 / 10

11 x 11 - 6 x 17 +4

8 / 10

5 x 50+57-57÷ 57

306

150

9 / 10

80+[20 x {20-(10÷5)}]

440

430

400

10 / 10

4+4+4+4÷ 4

Your score is

The average score is 49%

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TRIANGLE QUIZ

Created on September 26, 2021 By

Bina singh

Triangle Quizzes

1 / 5

Find the area of an isosceles triangle whose equal sides are 5 cm each and the base is 6 cm.

12 cm^{2}

6 cm^{2}

9 cm^{2}

15 cm^{2}

2 / 5

Find the area of an equilateral triangle whose side is $4\sqrt{3}$ cm.

12 sqrt{3} cm^{2}

14 sqrt{3} cm^{2}

115 sqrt{3} cm^{2}

30 sqrt{3} cm^{2}

3 / 5

The base of a right angled triangle is 8 cm and hypotenuse is 17 cm. Find its area.

15 square cm

21 square cm

18 square cm

60 square cm

4 / 5

Find the area of an equilateral triangle with side 10 cm.

25 sqrt(3) square cm

25 square cm

5 sqrt(3) square cm

20 sqrt(3) square cm

5 / 5

Find the area of the given triangle.

15 square cm

12 square cm

10 square cm

24 square cm

Your score is

The average score is 52%

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PERCENTAGE QUIZ

Percentage

NUMBER OF QUESTIONS 10

1 / 10

What % of 85 is 17 ?.

20%

40%

100

Let x% of 85 is 17 . Then $\frac{x}{100}\times 85=17$
$\Rightarrow 85x = 1700$
$\Rightarrow x= \frac{1700}{85}$
$\Rightarrow x= 20%$

2 / 10

........... % of 360 +15% of 820= 231. Find the missing number .

Let the missing number be x. Then x% of 360 +15% of 820= 231
$\Rightarrow \frac{x}{100}\times 360 + \frac{15}{100}\times 820 = 231$
$\Rightarrow 3.6x + 123 =231$
$\Rightarrow 3.6 x= 231-123 =108$
$\Rightarrow x= \frac{108}{3.6}= 30$
Hence the missing number is 30

3 / 10

If 40% of 140 = x% of 200, then find the value of x .

Given 40% of 140 = x% of 200
$\Rightarrow \frac{40}{100}\times 140= \frac{x}{100}\times 200$
$\Rightarrow 56 =2x$
So x= 28

4 / 10

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

588

600

672

700

Suppose originally he had x apples.

Then, (100 - 40)% of x = 420.

60% of x =420

60/100 x x =420

X=700

5 / 10

If 20% of a = b, then b% of 20 is the same as:

4% of a

5%of a

None of these

20% of a

6 / 10

The value of a washing machine depreciates at the rate of 10% every year. If its present value is Rs. 8748, then what was the price of the washing machine three years ago

12000

14000

60000

12500

Current price of the washing machine = Rs.8748

The price of the machine depreciated at the rate of 10% every year

Therefore, the price of the washing machine three years ago = 8748 ÷ (1 – 10/100)³

= Rs. [8748 x (10/9) x (10/9 ) x (10/9)]

= Rs.12000

7 / 10

If the price of a product is first decreased by 25% and then increased by 20%, then what is the percentage change in the price?

Let the original price be Rs. 100.

New final price = 120 % of (75 % of Rs. 100)

= Rs. [(120/100) x (75/100) x 100]

= Rs. 90

Therefore, the net change in price is 100 – 90 = 10.

Percentage decrease = 10%

8 / 10

A number is decreased by 10% and then increased by 10%. The number so obtained is 10 less than the original number. What was the original number?

3000

2000

1000

4000

Let the original number be x

Final number obtained = 110% of (90% of x)

=(110/100 × 90/100 × x)

= (99/100)x

Given the number obtained is 10 less than the original number.

x – (99/100) x = 10

Hence,

x =1000

9 / 10

What percentage of 2/7 is 1/35 ?

10 / 10

If 16% of 40% of a number is 8, then find the number.

120

125

300

450

Let X be the required number.

Therefore, as per the given question,

(16/100) × (40/100) × X = 8

So, X = (8 × 100 × 100) / (16 × 40)

= 125

Your score is

The average score is 70%

Ratio and Proportion Quiz

Ratio and proportions

1 / 10

The total number of workers at a factory is 120. There are 5 departments with 20, 30, 30, 20 and 20 employees, respectively. If the number of employees of the first two departments increase by 40% and the employees of the rest of the departments decrease by 60%, find the new ratio of employees.
a) 15 : 20 : 6 : 4 :
b) 15 : 22 : 6 : 6 : 4
c) 14 : 21 : 6 : 4 : 4
d) 14 : 21 : 6 : 6 : 4

15 : 20 : 6 : 4 : 4

15 : 22 : 6 : 6 : 4

14 : 21 : 6 : 4 : 4

14 : 21 : 6 : 6 : 4

Explanation: The new number of employees for the first department = 20 + 40% of 20 = 20 + 8 = 28
The new number of employees for the second department = 30 + 40% of 30 = 30 + 12 = 42
The new number of employees for the third department = 30 – 60% of 30 = 30 – 18 = 12
The new number of employees for the fourth department = 20 – 60% of 20 = 20 – 12 = 8
The new number of employees for the fifth department = 20 – 20% of 20 = 20 – 12 = 8
The new ratio = 28 : 42 : 12 : 8 : 8 = 14 : 21 : 6 : 4 : 4

2 / 10

The ratio of two numbers is 4 : 5. When 15 is added to the numbers the ratio changes to 11 : 13. Find the numbers.

40 , 50

32, 40

44, 55

36, 45

Explanation: Let the numbers be in terms of x.
The numbers are 4 : 5 = 4x : 5x
4x + 15 : 5x + 15 = 11 : 13
(4x + 15) / (5x + 15) = 11 / 13
52x + 195 = 55x + 165
3x = 195 – 165 = 30
3x = 30
X = 30 / 3 = 10
4x = 40, 5x = 50

3 / 10

The product of two numbers is 10 times their sum. Find the sum of the numbers if they are in a ratio 1 : 1.

200

400

Explanation: Let the number be in terms of x.
The sum = x + x = 2x
The product = x * x = x²
X² = 2x * 10 = 20x
The only possible value of x that satisfies the condition is 20.
So, the number is 20.
The sum = 20 + 20 = 40

4 / 10

The ratio of the measures ∠A and ∠B of a triangle ABC is 3 : 2. The ratio of the measures of ∠B and ∠C is 4 : 5. Find the measure of largest angle of the triangle ABC.

72°

78°

48°

60°

Explanation: : ∠ A : ∠ B = 3 : 2. ∠ B : ∠ C = 4 : 5. ∴ ∠ A : ∠ B : ∠ C = 6 : 4 : 5.
∴ actual values are 6x, 4x and 5x. So 6x + 4x + 5x = 180. This implies 15x = 180
i.e. x = 12. So angles are 6(12) = 72, 4(12) = 48 and 5(12) = 60

5 / 10

There are 145 students in the first three standards. The ratio of number of students in the first and the second standards is 2 : 3, while that of students in standards second and third is 4: 3. Find the number of students in 2^nd standard.

Explanation: : Total students = 145. Ratio of students in 1st and 2nd standards = 2 : 3
Ratio of students in 2^nd and 3^rd standards = 4 : 3
Hence combined ratio i.e. 1^st: 2^nd: 3^rd is 8 : 12 : 9.
∴ Number of students in each standard = (145 × 8)/29 = 40, (145 × 12)/29 = 60 and (145 × 9)/29 = 45.

6 / 10

The ratio of two numbers is 9 : 5. If 9 is added to the greater number and 5 is subtracted from the smaller number, the greater number becomes thrice the smaller one. Find the numbers.

72, 40

18, 10

36, 20

None of these

Present ratio = 9 : 5. ∴actual values are 9x and 5x.
Hence 9x + 9 : 5x – 5 = 3 : 1 ⇒ x = 4. ∴ numbers are 9(4) = 36 and 5(4) = 20.

7 / 10

If a: b = 5: 9 and b: c = 7: 4, then find a: b: c.

35:63:36

35:63:30

18:63:36

35:63:18

Here, we make the common term ‘b’ equal in both ratios.
Therefore, we multiply the first ratio by 7 and the second ratio by 9.
So, we have a : b = 35 : 63 and b : c = 63 : 36
Thus, a : b : c = 35 : 63 : 36

8 / 10

The fourth proportional to 5, 8, 15 is:

Let the fourth proportional to 5, 8, 15 be x.

Then, 5 : 8 : 15 : x

5x = (8 x 15)

X=(8x15)/5

X=24

9 / 10

A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?

500

1500

2000

none of these

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x - 3x = 1000

x = 1000.

B's share = Rs. 2x = Rs. (2 x 1000) = Rs. 2000.

10 / 10

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:

2:5

3:5

4:5

6:7

Let the third number be x.

Then, first number = 120% of x = 120x/100 = 6x/5

Second number = 150% of x = 150x/100 = 3x/2

Ratio of first two numbers = 6x/5 : 3x /2= 12x : 15x = 4 : 5

Your score is

The average score is 50%

AVERAGE QUIZ

Average quiz in hindi

Total number of questions: 10

Language : Hindi

1 / 10

एक कक्षा में 36 विद्यार्थीयों की औसत आयु 14 वर्ष है| जब एक शिक्षक को सम्मिलित किया जाता है तो औसत आयु 1 वर्ष बढ़ जाती है |शिक्षक की आयु क्या है?

36 विद्यार्थियों के आयु का योग = 36 x 14 = 504 वर्ष
36 विद्यार्थी और 1 शिक्षक की औसत आयु = 15 वर्ष
36 विद्यार्थी और 1 शिक्षक की औसत आयु का योग= 15 x 37= 555 वर्ष
शिक्षक की आयु = 555-504 =51 वर्ष

2 / 10

पांच संख्याओं का औसत 27 है| यदि उसमे से एक संख्या को हटा दिया जाये तो औसत 25 हो जाता है, तो निकाली गयी संख्या क्या है?

पांच संख्याओं का औसत= संख्याओ का योग/5
संख्याओं का योग = संख्याओं का औसत x 5
= 27 x 5
=135
माना निकाली गयी संख्या y है| तब चार संख्याओ का औसत = चार संख्याओं का योग /4
25= (135-y)/4
100= 135-y
y= 135-100
=35
अतः निकाली गयी संख्या 35 है|

3 / 10

3 वर्ष पहले पति, पत्नी और उनके बच्चे की औसत आयु 27 वर्ष थी और 5 वर्ष पहले पत्नी और बच्चे की औसत आयु 20 वर्ष थी। पति की वर्तमान आयु क्या है?

माना पति की वर्तमान आयु x वर्ष , माता की y वर्ष और बच्चे की वर्तमान आयु z वर्ष है|इसलिए 3 वर्ष पहले पति की वर्तमान आयु (x-3) वर्ष , माता की (y-3) वर्ष और बच्चे की वर्तमान आयु (z-3) वर्ष होगी| दिया गया है :
3वर्ष पहले पति, पत्नी और उनके बच्चे की औसत आयु = 27

$\frac{(x-3)+(y-3)+(z-3)}{3} =27$
$(x-3)+(y-3)+(z-3) =27 \times 3$
$x+y+z =81+9$
$x+y+z = 90 ........(i)$

5 वर्ष पहले पत्नी की आयु=(y-5 ) वर्ष
5 वर्ष पहले बच्चे की आयु=(z-5 ) वर्ष
5 वर्ष पहले पत्नी और बच्चे की औसत आयु = 20
$\frac{(y-5)+(z-5)}{2} =20$
$(y-5)+(z-5) =40$
इसलिए $y+z=50$ ...........(ii)
समीकरण (i) से (ii) को घटाने पर $x=90-50$
$x=40$
अतः पति की वर्तमान आयु 40 वर्ष है|

4 / 10

यदि P और Q की औसत मासिक आय रु. 5050 है, Q और R की औसत मासिक आय रु. 6250 और P और R की औसत मासिक आय रु. 5200 है तो P की मासिक आय क्या है?

4000

5000

6000

7000

माना P की मासिक आय Rs. x , Q की Rs. y और R की Rs. z है | दिया गया है : P और Q की औसत मासिक आय रु. 5050 है, Q और R की औसत मासिक आय रु. 6250 और P और R की औसत मासिक आय रु. 5200 है |
P और Q के मासिक आय का योग = 5050 x 2 = 10100
x+y= 10100 .......(1)
Q और R के मासिक आय का योग = 6250 x 2 =12500
y+z= 12500........(2)
P और R के मासिक आय का योग = 5200 x 2 =10400
x+z= 10400 .......(3)
समीकरण (1), (2) और (3) को जोड़ने पर 2(x+y+z)=33000
x+y+z= 16500 ...........(4)
समीकरण (4) में से समीकरण (2) को घटाने पर (x+y+z)-(y+z)=16500-12500
x= 4000
अतः P की मासिक आय = Rs. 4000

5 / 10

एक दुकानदार ने लगातार 5 महीनों तक 6435, रु. 6927, रु. 6855 रु. 7230 और रु. 6562 रु. कि बिक्री की | छठे महीने में उसकी कितनी बिक्री होनी चाहिए ताकि उसे औसतन बिक्री रु. 65०० हो सके?

4991

5000

4500

7500

माना छठे महीने में उसकी बिक्री x रु है| तब कुल महीनों का औसत

$औसत के सवाल$

$\frac{34009+x}{6}=6500$

$34009+x=39000$

$x=39000-34009$

$x=4991$

6 / 10

चार बहनों की औसत आयु 7 वर्ष है यदि माँ की आयु शामिल कर दी जाये तो औसत आयु 6 वर्ष बढ जाती है तो माँ की आयु होगी ?

दिया गया है : चार बहनों की औसत आयु =7
चार बहनों के आयु का योग/4=7
चार बहनों के आयु का योग = 4 x 7 = 28
माँ को शामिल करने पर औसत आयु = 7+6=13
माँ को शामिल करने पर आयु का योग = 13 x 5=65
माँ की आयु = माँ को शामिल करने पर आयु का योग- चार बहनों के आयु का योग
=65-28
=37

7 / 10

यदि 10 संख्याओं का औसत 24 है और 5 संख्याओं का औसत 15 है तो संयुक्त औसत ज्ञात कीजिये|