Quantitative Aptitude Quiz for Upcoming IBPS Exams – Set 22 | Word Problems

(1) The Cost Price of two car is same. One is sold at a profit of 20% and other for Rs 360 more than the first one. If the overall profit earned after selling the Car is 22%, then what is Cost Price of each car ?
A) 10,00,000.
B) 15,00,000.
C) 9,00,000.
D) Cannot be determined.
E) None of These

(2) Total number of questions in a exam is n. Mr. Sigmund gives right answer of 20 out 25 questions in first part. In second part he gives 60% right answer. In this way he scored 66.67% marks in exam.
If marks allotted to each question is equal and no negative marking is allowed. then total number of questions = ??

A) 85.
B) 75.
C) 90.
D) 50.
E) None of these

(3) A man sells two cycles for 32000 each one at a loss of 20% and other at a gain of 33.33% what is his overall  gain/loss ?
A) 20% profit.
B) 16% loss.
B) 4% loss.
D) no profit no loss.
E) None of these

(4) One house and one shop is being sold by a merchant. S.P of each is 1 lakh and on house he got 20% loss and on shop 20% profit. What is overall profit/loss in whole transaction ?
A) 1/12 lakh loss.
B) 3/14 lakh profit.
C) 12.5 lakh loss.  
D) 10 lakh profit.
E) None of these

(5) A dealer marks up price of oil by 20% uses faulty weigh of 900 gm in place of 1000 gm and gives a discount of 40% to his customer what is his loss percentage ?
A) 20%.
B) 35%.
C) 25%.
D) 28%.
E) None of these

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(6) Ram used a 140 cm instead of 1 mtr tape while buying clothe. While selling he used to 80 cm tape instead of 1 mtr tape. If he offers 20% discount while selling. What is his overall percentage gain?
A) 40%.
B) 35%.
C) 25%.
D) 60%.  
E) None of these

(7) The price of petrol increse from Rs 140/ltr to Rs 180/ltr by what % should cut down on his petrol consumption. If he can afford to to increase his expenditure on petrol by just 9.09% ?
A) 5.88%
B) 5.76%
C) 5.26%.
D) 3.79%.
E) None of these

(8)A, B and C can do a work in 24, 30 and 40 days respectively. If they work such that A works for 1 day, then B for 2 days, then C for 3 days. Then again A for 1 day and so on. How many day will it take to complete the work ?
A) 31.75 days.
B)32.25 days.
C) 32.33 days.
D)32.67 days.
E) None of these

Answers:

Solutions :-

1. C
2. D
3. D
4. A
5. A
6. A
7. E
8. B

1) 9,00,000
 Average % profit = 22
 So, total % profit = 22×2 = 44%
 44 – 40 = 4% (40 arrived after 20×2)
 4% = 360
 100% = 9,00,000

2) 75

20/25 = 4/5 = 80% (ri8 in first part)

80%                60%
   
            66.67%

20/3             40/3

1:2

1= 25
So, 3 = 75 and.

3) no profit no loss

Two case occurred here one is loss of 20% and other is profit of 33.33%.
( I try to solve this by reciprocal method which is also useful in further questions)

Case 1) Loss 20%
So it sold at 4/5

Case 2) profit 33.33%
So it sold at 4/3

In this type of questions S.P is supposed to L.C.M of numenator of both reciprocal…

So, in Case1)
S.P = 4
C.P= 4*5/4 = 5

Similarly in case 2)
S.P = 4
C.P = 4*3/4= 3

Total S.P = 4+4 = 8
Total C.P = 5+3= 8

So no profit no loss.

(It looks long in explanation but believe me it’s very short and conceptual way to get right answer quickly. It’s not trick. It’s concept.)
           

4) 1/12 lakh loss
 same as question no 3

Case 1)
Loss 20%
It sold at 4/5

Case 2)
Profit 20%
It sold at 6/5

S.P in case (1) = 12 (LCM of both numenator)
C.P = 12*5/4 = 15

S.P in case (2) = 12
C.P = 12*5/6 = 10

Total S.P = 24
 C.P = 25

When each S.P is 12,  loss is Rs 1
So when each S.P is 1 lakh, loss is 1/12 lakh

5) 20% loss
In this types of questions always remember SP/CP.

120/100 ( he tried to get 20% profit while buying)

1000/900 ( he used to 900gm instead of 1000gm it means he paid C.P of 900gm but take money of 1000gm)

60/100 ( he gave 40% discount)

Multiply all three to get final reciprocal which is 4/5.
It means loss of 1/5 = 20%

6) 40% profit

Same rule as question no 5
SP/CP =
140/100
*100/80
*80/100

= 7/5
Profit % = 2/5= 40 %

7) none of these

Expenditure = price × Consumption
12/11 = 180/140  * X
X = 28/33
It means reduction = 5/33%

8) 32.25 days

Total work is 120 unit ( LCM of 24, 30 and 40)
A = 24 days ( 5 unit@day) works for 1 day =  5 unit
B = 30 days ( 4 unit@day) works for 2 days = 8 unit
C = 40 days ( 3 unit@day) works for 3 days = 9 unit

6 days = 22 units
5×6days = 22×5 units
30 days = 110 units
31st day = 115 units ( 5 units by A)
32nd day =119 units ( 4 units by B)
Remaining 1 unit will being done by B in 1/4 day

So total day is 32.25 days.

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