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Question 1 of 10
1. Question
In an examination, there are three papers and a candidate has to get 40% of the total to pass. In one paper, he gets 62 out of 150 and in the second 45 out of 150. How much must he get, out of 180, in the third paper to just qualify for a pass?
Correct
Let the marks required be x. Then, (62 + 45 + x) = 40% of (150 + 150 + 180)
⇔ 107 + x = (40/100) ×480 ⇔ x=192-107=85.
Incorrect
Let the marks required be x. Then, (62 + 45 + x) = 40% of (150 + 150 + 180)
⇔ 107 + x = (40/100) ×480 ⇔ x=192-107=85.
Question 2 of 10
2. Question
At what percent above the cost price must a shopkeeper mark his goods so that he gains 40% even after giving a discount of 20% on the marked price?
Correct
Let C.P. = Rs. 100. Then, S.P. = Rs. 140.
Let marked price be Rs. x. Then, 80% of x = 140 ⇒ x= ((140 × 100)/80) =175
∴ Marked price =75% above C.P.
Incorrect
Let C.P. = Rs. 100. Then, S.P. = Rs. 140.
Let marked price be Rs. x. Then, 80% of x = 140 ⇒ x= ((140 × 100)/80) =175
∴ Marked price =75% above C.P.
Question 3 of 10
3. Question
The prices of a scooter and a T.V. are in the ratio 6: 5. If the scooter costs Rs. 10000 more than a T.V. set, then the price of a T.V. set is:
Correct
Let the prices of a scooter and a T.V. set be Rs. 6x and Rs. 5x respectively. Then, 6x – 5x = 10000 ⇔ x = 10000
∴ Price of T.V. set = Rs. (5 × 10000) = Rs. 50000.
Incorrect
Let the prices of a scooter and a T.V. set be Rs. 6x and Rs. 5x respectively. Then, 6x – 5x = 10000 ⇔ x = 10000
∴ Price of T.V. set = Rs. (5 × 10000) = Rs. 50000.
Question 4 of 10
4. Question
The sum of five consecutive even numbers is equal to 180. What is the sum of the largest number amongst them and the square of the second smallest number amongst them together?
Correct
Let the five consecutive even numbers are 2x, 2x + 2, 2x + 4, 2x + 6, 2x + 8 respectively.
According to the question,
2x + 2x + 2 + 2x + 4 + 2x + 6 + 2x + 8 = 180
10x + 20 = 180
10x = 160
x=160/10=16
∴ The five numbers are 32, 34, 36, 38, 40 respectively.
∴ Required sum =(34)^{2}+40=1156+40=1196
Incorrect
Let the five consecutive even numbers are 2x, 2x + 2, 2x + 4, 2x + 6, 2x + 8 respectively.
According to the question,
2x + 2x + 2 + 2x + 4 + 2x + 6 + 2x + 8 = 180
10x + 20 = 180
10x = 160
x=160/10=16
∴ The five numbers are 32, 34, 36, 38, 40 respectively.
∴ Required sum =(34)^{2}+40=1156+40=1196
Question 5 of 10
5. Question
The size of a plastic block is 15*30 *60 cm. how many such blocks will be required to construct a solid plastic cube of minimum size?
Correct
Side of smallest cube = LCM of 15, 30 and 60 LCM of 15, 30 and 60 = 60cm Volume of the cube = (60*60*60) cu.cm =216000cu.cm Volume of the block = (15*30*60) cu.cm =27000 cu.cm No. of blocks = volume of the cube / volume of the block =216000 / 27000 =8
Incorrect
Side of smallest cube = LCM of 15, 30 and 60 LCM of 15, 30 and 60 = 60cm Volume of the cube = (60*60*60) cu.cm =216000cu.cm Volume of the block = (15*30*60) cu.cm =27000 cu.cm No. of blocks = volume of the cube / volume of the block =216000 / 27000 =8
Question 6 of 10
6. Question
Manohar got 56 marks in Hindi, 75 marks in Science, 89 marks in Maths, 69 marks in Social Science and 91 marks in English. The maximum marks of each subject are 100. How much overall percentage of marks did he get?
Correct
Marks in all subjects = 56 + 75 + 89 + 69 + 91 = 380
Total maximum marks = 5 × 100 = 500
∴ Required percentage =380/500×100=76%
Incorrect
Marks in all subjects = 56 + 75 + 89 + 69 + 91 = 380
Total maximum marks = 5 × 100 = 500
∴ Required percentage =380/500×100=76%
Question 7 of 10
7. Question
7 women cam complete a piece of work in 16 h. In how many hours will 14 women complete the same piece of work?
Correct
7 women can complete a piece of work in = 16 h
1 woman can complete a piece of work in = (16 × 7) h = 112 h
∴ 12 women can complete a piece of work in =112/14 = 8 h
Incorrect
7 women can complete a piece of work in = 16 h
1 woman can complete a piece of work in = (16 × 7) h = 112 h
∴ 12 women can complete a piece of work in =112/14 = 8 h
Question 8 of 10
8. Question
The length of a rectangle is 42 cm which is 14 cm more than the diameter of a circle. What is the area of the circle?
Correct
∴ Diameter of a circle = 42 – 14 = 28 cm
∴ Radius of a circle =28/2 = 14 cm
∴ Area of a circle =πr^2=22/7×14×14=616 sq cm
Incorrect
∴ Diameter of a circle = 42 – 14 = 28 cm
∴ Radius of a circle =28/2 = 14 cm
∴ Area of a circle =πr^2=22/7×14×14=616 sq cm
Question 9 of 10
9. Question
In a business partnership, P invests 1/3 of the capital for 1/3 of the time; q invests ½ of the capital for ½ of the time & R, the rest of the capital for the whole time. Out of a profit of Rs.7600 Q’s share is.
Correct
Suppose P invests Rs x/3 for y/3 months Then Q invests Rs x/2 for y/2 months R invests [x – (x/3+x/2)] for y months R invests [x – ((2x+3x)/6)] for y months R invests [x – 5x/6] for y months R invests x/6 for y months Therefore P:Q:R = (x/3*y/3) : (x/2*y/2) : (x/6*y) =xy/9:xy/4:xy/6 =1/9:1/4:1/6 =4:9:6 Hence Q ‘s share = Rs(7600*9/19) =Rs.3600
Incorrect
Suppose P invests Rs x/3 for y/3 months Then Q invests Rs x/2 for y/2 months R invests [x – (x/3+x/2)] for y months R invests [x – ((2x+3x)/6)] for y months R invests [x – 5x/6] for y months R invests x/6 for y months Therefore P:Q:R = (x/3*y/3) : (x/2*y/2) : (x/6*y) =xy/9:xy/4:xy/6 =1/9:1/4:1/6 =4:9:6 Hence Q ‘s share = Rs(7600*9/19) =Rs.3600
Question 10 of 10
10. Question
A certain newspaper is sold at a certain price. By selling it at 4/5 of that price one loss 5%. Find the gain percent at original cost.
Correct
Let the original selling price be Rs.x Loss = 5% New SP = 4x/5 So cost price = Rs (100/95 * 4x/5) =Rs. (16x / 19) Now cost price = Rs.16x / 19 Selling price = Rs.x Gain = Rs(x – 16x /19) =Rs.3x / 19 Gain% = gain*100 / C.P =3x*100*19 / 19*16x =300 / 16 =18.75%
Incorrect
Let the original selling price be Rs.x Loss = 5% New SP = 4x/5 So cost price = Rs (100/95 * 4x/5) =Rs. (16x / 19) Now cost price = Rs.16x / 19 Selling price = Rs.x Gain = Rs(x – 16x /19) =Rs.3x / 19 Gain% = gain*100 / C.P =3x*100*19 / 19*16x =300 / 16 =18.75%