M4maths Previous Puzzles

13 Dec 2021 Sum

The sum of a number of two digits and the number obtained by exchanging the places of digits is 110. If units place is 4 more than the tens place in the given number, then product of two digits of the number will be

OPtion

1) 64
2) 46
3) 37
4) 73
5) 19
6) 91
7) 82
8) 28
9) 21
10) none of these

Solution

let no be yx
given x-y=4...(1)
xy+yx=110 equivalent to x+y+10(x+y)=110
i.e. x+y=10...(2)
solving (1)&(2) we get x=7,y=3
i.e. the no. is 37
product will be 3*7=21

Correct Option: 9 Best Solution (3)

10 Dec 2021 Length of bridge

A train 130 m long passes a bridge in 21 seconds moving at a speed of 90 Km/h. Find out the length of the bridge

OPtion

1. 395 m.
2. 420 m.
3. 425 m.
4. 380 m.
5. 386 m.
6. 352 m.
7. 365 m.
8. 368 m.
9. 399 m.
10. None of the above

Solution

We have, speed of the train
= (length of the train + length of the bridge) / time taken in crossing the bridge
⇒ 5/18 x 90 = (130 + length of the bridge)/ 21
= 450 x 21 = 2340 + 18 (length of the bridge)
= 9450 – 2340 = 18 (length of the bridge)
∴ Length of the bridge
= 7110 / 18 = 395 m.

Correct Option: 1 Best Solution (6)

09 Dec 2021 Class

The average weight of 31 students of section A of a class is 53 Kg, whereas the average weight of 27 students of section B of the same class is 55 Kg. Find out average weight of all the 58 students of the class

OPtion

1. 59.3 kg
2. 55.5 kg
3. 52.6 kg
4. 53.9 kg
5. 55.6 kg
6. 58.2 kg
7. 54.2 kg
8. 53.1 kg
9. 60.2 kg
10. None of the above

Solution

Here, n1 = 31, n2 = 27, x = 53, and y = 55.
∴ Average weight of all the 58 students = (n1x + n2y)/(n1+n2)
= [(31x53) +(27x55)]/ (31+27)
= (1643 +1485)/58
= 53.9kg

Correct Option: 4 Best Solution (3)

08 Dec 2021 AVERAGE

The average of x numbers is 5x. If x − 2 is subtracted from each given number, what will be the new average?

OPtion

1. 2X + 2
2. 2X + 1
3. 2X - 1
4. 2X – 2
5. 4X + 1
6. 4X + 2
7. 4X – 1
8. 5X + 2
9. 5X + 1
10. None of the above

Solution

The new average = x − a
=5x − (x − 2)
= 4x + 2.

Correct Option: 6 Best Solution (1)

07 Dec 2021 cistern

A pipe can empty a cistern in 40 minutes. Find the time in which 3/4 part of the cistern will be emptied.

OPtion

1. 25 minutes
2. 28 minutes
3. 32 minutes
4. 40 minutes
5. 30 minutes
6. 36 minutes
7. 38 minutes
8. 32 minutes
9. 29 minutes
10. None of the above

Solution

1/40 part of the cistern is emptied in = 1 minute.
∴ 3/4 part of the cistern is emptied in
= 40 × 3/4 = 30 minutes.

Correct Option: 5 Best Solution (2)

06 Dec 2021 Gain

A shopkeeper professes to sell his goods on cost price, but uses 750 gm, instead of 1.2 Kg. What is his gain %?

OPtion

1. 25%
2. 30%
3. 33.33%
4. 20%
5. 40%
6. 55%
7. 50%
8. 60%
9. 52%
10. None of the above

Solution

Here, cost price of 1200 gm is equal to selling price of 750 gm,
∴ % gain = [(m – n)/ n] × 100
= [(1200 – 750)/ 100] x100
= 450/ 750 × 100 = 60%

Correct Option: 8 Best Solution (4)

03 Dec 2021 integers

Arrange 9 integers from 1 to 9 without repetition in a 3*3 size square box (total 9 small boxes) in such a manner that sum of 3 integers (S) of 3 horizontal , 3 vertical and 2 diagonal lines are same. S will be

OPtion

1) 9
2) 10
3) 12
4) 14
5) 15
6) 18
7) 21
8) 25
9) 28
10) None of the above

Solution

sum of all 9 integers = 45
so sum S = 45/3 = 15

Correct Option: 5 Best Solution (0)

02 Dec 2021 Sum of age

A father said to his son that his (father) age was 7 times his (son) age and his mothers age was 6 times his age 15 years before. If age of father is 5 years more than that of mother, then sum of their ages (all the three) at present will be

OPtion

1) 105
2) 125
3) 130
4) 100
5) 95
6) 90
7) 85
8) 115
9) 80
10) none of these

Solution

let age of father = x
that of mother = x-5
and that of son = y
now according to the given conditions
x-15 = 7(y-15) or x-7y = -90
x-5-15 = 6(y-15) or x-6y = -70
on solving we will get y = 20 and x = 50
so sum = 50 + 45 + 20 = 115

Correct Option: 8 Best Solution (0)

01 Dec 2021 Discount

find the true discount of ₹6300 due in 4 years at 12.5% per annum at simple interest.

OPtion

1. ₹2100
2. ₹2350
3. ₹2700
4. ₹2500
5. ₹2550
6. ₹2210
7. ₹2360
8. ₹2110
9. ₹2160
10. None of the above

Solution

T.D. = Amount - P.W.
P.W.= [(100 x 6300)/(100 + 12.5 x 4)]
P.W. = (630000 / 150)
P.W. = ₹ 4,581.81

T.D. = ₹6300 - ₹ 4,200
= ₹2100

Correct Option: 1 Best Solution (1)

30 Nov 2021 Investment

Find out the investment required to purchase ₹75000 of 10% stock at 95.

OPtion

1. ₹70000
2. ₹72500
3. ₹72250
4. ₹70450
5. ₹71500
6. ₹71250
7. ₹71350
8. ₹74000
9. ₹73580
10. None of the above

Solution

Market value of ₹100 stock = ₹95
∴ Market value of ₹75000 stock = ₹ {(95/100) ×75000}
= ₹71250

Correct Option: 6 Best Solution (0)