M4maths Previous Puzzles

22 Jul 2022 Sequence & Series

Which one set of letters when placed sequentially in the blanks will complete the following series correctly ?
a _ c a _ b _ a _ _ b c _ a b _ c c _ a a b _ c _

OPtion

1) babacababc
2) bacabbcabc
3) bcbabababc
4) bacabaabcc
5) babacbbabc
6) bacabababc
7) bacbcababc
8) bacacacabc
9) babacababc
10) None of these

Solution

The series is abc, aabc, aabbc, aabbcc, aaabbcc, ...
So letters in the blank spaces are: bacabababc
Correct Option 6)

Correct Option: 6 Best Solution (2)

21 Jul 2022 Time Distance & Speed

A and B take part in a 500 m race. A runs at 24 km per hour. A gives B a start of 40 m and still beats him by 15 seconds. The speed of B is

OPtion

1) 15.5 km/hr
2) 16.4 km/hr
3) 14.5 km/hr
4) 16.8 km/hr
5) 18.4 km/hr
6) 15.2 km/hr
7) 20.5 km/hr
8) 17.4 km/hr
9) 15.8 km/hr
10) None of these

Solution

A's speed = 24 x (5/18) m/s = 20/3 m/s
Time taken by A to cover 500 m = 500/(20/3) = 75 sec
.'. Time taken by B to cover 460 m = (75+15) = 90 sec.
Hence, B's speed = (460/90) x (18/5) = 92/5 m/s = 18.4 km/h
Correct Option 5)

Correct Option: 5 Best Solution (2)

20 Jul 2022 Probability

A bag contains 5 Blue, 6 Green, 3 Yellow and 4 Red balls. If four balls are selected at random, then what is the probability that there exist at least two red balls ?

OPtion

1) 2/9
2) 1/20
3) 301/1530
4) 134/765
5) 67/170
6) 1/204
7) 67/340
8) 57/340
9) 134/340
10) None of these

Solution

Total ways of selecting 4 balls out of 18 = 18C4 = 3060
Out of the 4 balls at least 2 red colour is possible when, there are (2 Red+2 Non-Red) or (3 Red + 1 Non-Red) Or All Four Red
.'. Ways of selecting at least 2 red balls = (4C2 x 14C2) + (4C3 x 14C1) + 4C4 = 546 + 56 + 1 = 603
Hence, required probability = 603/3060 = 67/340
Correct Option 7)

Correct Option: 7 Best Solution (2)

19 Jul 2022 Permutation & Combination

In a school, a student can opt the language subject out of English, Hindi and Sanskrit either single or in a combination. 50 students opted English, 62 Hindi and 42 Sanskrit. If there are total 78 students and only 12 opted for all three languages, then the number of students who opted for exactly two languages is

OPtion

1) 72
2) 62
3) 76
4) 52
5) 45
6) 80
7) 50
8) 48
9) 60
10) None of these

Solution

Let E, H, S be the sets of students who opted for English, Hindi and Sanskrit respectively.
n(E)=50
n(H)=62
n(S)=42
n(E∪H∪S)=78
n(F∩B∩C)=12
We know, n(E∪H∪S)=n(E)+n(H)+n(S)−n(E∩H)−n(H∩S)−n(E∩S)+n(E∩H∩S)
78 = 50 + 62 + 42 −n(E∩H)−n(H∩S)−n(E∩S) + 12
⇒ n(E∩H) + n(H∩S) + n(E∩S) = 88
"The number of students who opted for exactly two languages = n(E∩H) + n(H∩S) + n(E∩S) − 3×n(E∩H∩S) = 88 − 3x12 = 52

Correct Option: 4 Best Solution (1)

18 Jul 2022 Average

The average weight of a group of boys is 48 kg. If 8 boys of average weight 51 kg leave and 3 boys of average weight 42 kg join the group, then the average weight of group decreases by 1 kg 50 gm. Find the initial number of boys in a group

OPtion

1) 38
2) 48
3) 42
4) 39
5) 45
6) 33
7) 49
8) 54
9) 40
10) None of these

Solution

Let there are 'n' boys in a group initially, then their total weight=48n
When, 8 boys of average weight 51 kg leave and 3 boys of average weight 42 kg join the group, then of boys left = n-8+3 = n-5
New average of a group = (48n - 8*51 + 3*42) / (n-5) = (48 - 1.05)
48n - 408 + 126 = 46.95n - 234.75
1.05n = 47.25
n = 45
Correct Option 5)

Correct Option: 5 Best Solution (1)

15 Jul 2022 Profit & Loss

A retailer sells two items having cost prices Rs. 1200 and Rs. 1500 respectively. He offers discount of 8% on both items. The ratio between the marked price of these items is 8:9 respectively. If the total profit earned by a retailer is Rs. 428, then the difference between the marked price (in Rs) of two items is

OPtion

1) 450
2) 500
3) 300
4) 350
5) 250
6) 100
7) 400
8) 550
9) 200
10) None of these

Solution

Given, CP are Rs. 1200 and Rs. 1500
Let the MP be 8x and 9x respectively.
Then with 8% discounts, MP are (92/100)*8x and (92/100)*9x
.'. Profits are (92/100)*8x - 1200 and (92/100)*9x - 1500
Given, sum of profits = 428
⇒ (92/100)*8x - 1200 + (92/100)*9x - 1500 = 428
(23/25)*17x = 3128
x = 200
Hence, difference of marked price = 9x - 8x = x = 200
Correct Option 9)

Correct Option: 9 Best Solution (1)

14 Jul 2022 Sequence & Series

What is the value of 'k' in the following sum ?
57 + 54 + 51+ ........+ k = -123

OPtion

1) 60
2) 63
3) -60
4) -51
5) -75
6) -63
7) -48
8) -54
9) -66
10) None of these

Solution

Given, arithmetic sequence 57, 54, 51, ....., k. Here first term a=57, Difference d=-3, Last term=k, Sum of all terms=-123
Sum of the n terms of an A.P. = (n/2)[2a + (n-1)d]
-123 = (n/2)[2*57 + (n-1)*(-3)]
⇒ n² - 39n - 82 = 0
(n-41)(n+2) = 0
Considering positive value of number of terms, n=41 ⇒ k=41th term of the series.
Now, n th term of an AP = a + (n-1)d
.'. k = 57 + 40*(-3) = -63

Correct Option: 6 Best Solution (0)

13 Jul 2022 Algebra

If, 3^(2x+y)=243 and 2^(x+2y)=256, then 6x+3y = ?

OPtion

1) 13/3
2) 11/3
3) 2/3
4) 12
5) 15
6) 18
7) 24
8) 30
9) 27
10) None of these

Solution

Given, 3^(2x+y)=243
⇒ 3^(2x+y) = 3^5
Comparing powers of a equal base: 2x+y = 5 ---(i)
Similarly, 2^(x+2y)=256
⇒ 2^(x+2y) = 2^8
x+2y = 8 ---(ii)
Solving, eqn. (i) & (ii), we get, x=2/3, y=11/3
.'. 6x+3y = 6*(2/3) + 3*(11/3) = 15
Correct Option 5)

Correct Option: 5 Best Solution (2)

12 Jul 2022 Ratio & Proportion

A number of light bulbs were purchased to illuminate a house. However, only 3/5 of them were needed and 72 leftover light bulbs were returned. If only 70% of the cost of those returned, or Rs. 11088, was reimbursed, how much money was spent on illuminating the house?

OPtion

1) Rs. 23760
2) Rs. 16632
3) Rs. 31680
4) Rs. 21384
5) Rs. 33264
6) Rs. 14256
7) Rs. 22176
8) Rs. 28512
9) Rs. 27720
10) None of these

Solution

Bulbs used=3/5 .'. Bulbs returned=2/5
Let total bulbs=x, then (2/5)*x=72 ⇒ x=180
We know that Rs. 11088 represents 70% of the cost of 72 bulbs
.'. Full cost of 72 bulbs=(100/70)*11088=15840
Cost per bulb=15840/72=220
And cost which was not recovered=15840-11088=4752
Total cost = Cost(108 used bulbs) + Cost not recovered = 108*220 + 4752 = 28512 Rs.

Correct Option: 8 Best Solution (1)

11 Jul 2022 Area and Volume

It cost Rs. 1089 to paint all the surfaces of a cube at the rate of Rs. 1.5 per meter sq. If the same cube is cut to form three cuboids of equal volumes, then the cost of painting all the surfaces of any two of the cuboids so formed at the 50% more rate is

OPtion

1) Rs. 2722.50
2) Rs. 1270.50
3) Rs. 1512.50
4) Rs. 1210
5) Rs. 1815
6) Rs. 2541
7) Rs. 3630
8) Rs. 1694
9) Rs. 2420
10) None of these

Solution

Surface area of a cube = 6a², where a=side
Cost of painting = Surface Area*Rate per meter sq.
1089 = 6a²*1.5
⇒ a²=121 or a=11
When the cube is cut to form three cuboids of equal volume, one of the edge (side) will be divided into 3 equal parts.
Then, sides of each cuboid are 11/3, 11 and 11 m.
.'. Surface area of a cuboid = 2*[(11/3)*11 + (11/3)*11 + 11*11] = 1210/3
New cost of painting = (150/100)*1.5 = 9/4/m sq.
Hence cost of painting 2 cuboids = 2*(9/4)*(1210/3) = 1815
Correct Option 5)

Correct Option: 5 Best Solution (1)