M4maths Previous Puzzles

15 Sep 2020 Percentage

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

OPtion

1) 10% of a
2) 2% of a
3) 16% of a
4) 40% of a
5) 8% of a
6) 4% of a
7) 8% of a
8) 6% of a
9) 16% of a
10) None of these

Solution

20% of a = b

=> (20/100) * a = b

b% of 20 =(b/100) x 20 = [(20a/100) / 100] x 20= 4a/100 = 4% of a.

Correct Option: 6 Best Solution (0)

14 Sep 2020 Cost Price

If the cost price of a Mobile is first decreased by 25% and then increased by 20%, then the net change in the price will be :

OPtion

1) 9%
2) 5%
3) 12%
4) 11%
5) 12.5%
6) 6%
7) 8%
8) 7%
9) 10%
10) None of these

Solution

Let the original price be Rs. 100.

New final price = 120 % of (75 % of Rs. 100) = Rs. [(120/100) * (75/100) * 100] = Rs. 90.

Decrease = 10%

Correct Option: 9 Best Solution (6)

11 Sep 2020 Arithmetic

Aflatoon, on his death bed, keeps half his property for this wife and divides the rest equally among his three sons : Ved, Charl and Dev. Some years later, Ved dies leaving half his property to his widow and half to his brothers Charl and Dev together, sharing equally. When Charl makes his will, he keeps half his property for his widow and the rest he bequeaths to his younger brother Dev. When Dev dies some years later, he keeps half his property for his widow and the remaining for his mother. The mother now has Rs. 1,827,000.What was the worth of the total property?

OPtion

1) 2480000
2) 2400000
3) 2460000
4) 2490000
5) 2784000
6) 2784800
7) 2784600
8) 2784200
9) 2784400
10) None of these

Solution

Let the total property of the Aflatoon be Rs.x.

After Aflatoon’s death, money possessed by the family members would be

Wife = x/2, Ved = x/6, Charl =x/6, Dev =x/6

After Ved’s death, money possessed by each of them would be

Aflatoon’s wife = x/2, Ved = 0,Ved’s wife = x/6, Charl = x/6 + x/24 = 5x/24, Dev = x/6 + x/24 = 5x/24

After Charl’s death, money possessed by them

Aflatoon’s wife has x/2, Ved has 0, Ved’s wife has x/12, Charl has 0, Charl’s wife has 5x/48, Dev has

5x/24 + 5x/48 = 15x/48

After Dev’s death, money possessed by them is:

Aflatoon’s wife has x/2 + 15x/96 = 63x/96, Ved has 0, Ved’s wife has x/12, Charl’s has 0, Charl’s wife has

5x/48, Dev has 0 and Dev’s wife has 15x/96

Now, given that 63x/96 = 1827000

x= 2784000

Correct Option: 5 Best Solution (4)

10 Sep 2020 Profit & Loss

A battery dealer incurs an expense of Rs. 150 for producing every battery. He also incurs an additional expenditure of Rs. 30,000, which is independent of the number of batteries produced. If he is able to sell a battery during the season, he sells it for Rs. 250. If he fails to do so, he has to sell each battery for Rs. 100.If he produces 1500 batteries, what is the number of batteries that he must sell during the season in order to break-even, given that he is able to sell all the batteries produced?

OPtion

1) 780
2) 700
3) 760
4) 800
5) 909
6) 1000
7) 900
8) 1100
9) 880
10) None of these

Solution

Total cost to produce 1500 battery = (1500 × 150 + 30000) = Rs. 2,55,000

Let he sells x batteries during the season, therefore

number of batteries sold after the season = (1500 – x)

∴Revenue earned on the sale of 1500 watches

= 250 × x + (1500 – x) × 100 = 150x + 150000

Now, break-even is achieved if production cost is equal to the selling price.

∴ 150x + 150000 = 2,55,000 ⇒x = 700

Correct Option: 2 Best Solution (4)

09 Sep 2020 Profit & Loss

A stockist wants to make some profit by selling rice. He contemplates about various methods. Which of the following would maximize his profit?
1. Sell rice at 10% profit.
2. Use 900 g of weight instead of 1 kg.
3. Mix 10% impurities in rice and selling rice at cost price.
4. Increase the price by 5% and reduce weights by 5%.

OPtion

1) 1
2) 2
3) 3
4) 4
5) 1 or 2
6) 2 or 3
7) 1 or 3
8) 1 or 4
9) 2 or 4
10) None of these

Solution

We will solve this question by taking all the cases one by one.

In the first case it is given that the profit is 10%.

For second case, let the CP of 1 kg of rice be Rs. 100

Then CP of 900 g of rice= (100/1000 )x 900 = Rs. 90

Hence, profit % in Case II= [{(100-90)/90}x100] = 11.11%

For case III, If he adds 10% impurity then his CP for 1 kg

= {(100/1100) x 100} = Rs. 90.90

Hence, profit % in Case III = [{(100-90.90)/90.90} x 100] = 10.01%

and in the last case, If he reduces weight by 5%

Then cost price of 950 g = {(100/1000) x 950} = Rs. 95 and SP = Rs. 105

Hence, profit % in Case IV = {(105 – 95 )/95} X 100 = 10.52%

Hence, the profit is maximum in second case.

Correct Option: 2 Best Solution (0)

08 Sep 2020 Cost Price

If 2 cricket bats and 3 balls cost Rs, 3500 and 3 cricket bats and 2 balls cost Rs. 4000, then how much does a cricket bat cost ?

OPtion

1) 800
2) 500
3) 600
4) 840
5) 640
6) 540
7) 480
8) 700
9) 1000
10) None of these

Solution

Let the cost of a bat and that of a ball be Rs. x and Rs, y respectively.

Then, 2x + 3y = 3500 ...(i)

and 3x + 2y = 4000 .....(ii)

solving (i) and (ii) we get x = 1000, y = 500

Correct Option: 9 Best Solution (0)

07 Sep 2020 Profit & Loss

Find the cost price of 1KG of Saffron which is sold for Rs. 58000 at a loss of 13%.

OPtion

1) 68000
2) 60000
3) 66000
4) 63000
5) 64000
6) 66661.67
7) 66663.67
8) 66666.67
9) 66669.67
10) None of these

Solution

If CP 100 SP 87
i.e. When SP 87, CP 100
when SP 58000, CP = 100*58000/87 = ~66666.67

Correct Option: 8 Best Solution (7)

04 Sep 2020 Arithmetic

Ms. Shakuntla asked his students to find the sum of all pages of their maths book. Pappu gave the answer as 1098. Ms. Shakuntla told him that he has not added one page.
Find the page number which the Pappu forgot to add?

OPtion

1) 27
2) 19
3) 33
4) 39
5) 38
6) 45
7) 30
8) 41
9) 43
10) None of these

Solution

1 + 2 + 3 + .........+n = 1098
⇒ n(n+1)/2 = 1098
⇒ n(n + 1) = 2196
⇒ n^2 = 2196.
Now, the perfect square closest 2196 is 2209, which is the square of 47.

Sum will be -
47*48/2 = 1128

As the given sum is 1128, so the Pappu forgot to add page number 1128 - 1098 = 30

Correct Option: 7 Best Solution (3)

03 Sep 2020 Permutation and Combination

How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at the unit’s place must be greater than that in the ten’s place?

OPtion

1) 44
2) 55
3) 60
4) 64
5) 63
6) 56
7) 66
8) 58
9) 72
10) None of these

Solution

1 at units place 0 numbers
2 at units 1 at units place 3!
3 at units 1 at units place 3!
3 at units 2 at units place 3!
4 at units 1 at units place 3!
4 at units 2 at units place 3!
4 at units 3 at units place 3!
5 at units 1 at units place 3!
5 at units 2 at units place 3!
5 at units 3 at units place 3!
5 at units 4 at units place 3!

Correct Option: 3 Best Solution (0)

02 Sep 2020 Arithmetic

Gabbar fires 5 shots to Mogambo's 3 but Gabbar kills only once in 3 shots while Mogambo kills once in 2 shots. When Mogambo has missed 27 times, Gabbar has killed :

OPtion

1) 51
2) 57
3) 42
4) 30
5) 49
6) 56
7) 33
8) 36
9) 32
10) None of these

Solution

Let the no of shots be x. Then,

Shots fired by Gabbar =(5/8)x

Shots fired by Mogambo = (3/8)x

Killing shots by Gabbar = 1/3 of (5/8)x = (5/24)x

Shots missed by Mogambo = 1/2 of (3/8)x = (3/16)x

(3/16)x = 27 ⇒ x =144

Killed by Mogambo = (5/24)x = (5/24)*144 = 30

Correct Option: 4 Best Solution (3)