Various quant questions are coming in IBPS Clerical exams. Questions are a bit calculative but all questions entertains the simple basic concepts. Here we’re providing the methods and approach to solve these questions.
Ques 1: The difference between 75% and 4/7 of a number is 125. Find the number.
This question is solved simply.
Let the number be X
Then 75% of X becomes 3X/4
Now, 3X/4 - 4X/7 = 125
Cross multiply → 5X/28 = 125 → X = 25*28 = 700
Ques 2: An amount is invested in scheme A for 6 years at simple interest at the rate of interest was 18.5%. If 3/4th of same amount is invested for 4 years in scheme B on simple interest, it becomes equal to the interest obtained in scheme A. What is the rate of interest in scheme B?
This question is based upon the basic concepts of simple interest.
SI = PRT/100
Now, let the amount invested (Principal) in Scheme A be P Rs.
According to question, SI in Scheme A = P*18.5*6/100 = 111P/100
Now, in second case, the amount invested in Scheme B is 3/4 of P = 3P/4
Now, SI on 3P/4 for 4 years will be (taking rate of interest as R%) = ([3P/4]*R*4)/100 = 3PR/100
This is SI, then final amount after we add SI to 3P/4 = 3PR/100 + 3P/4
This amount is equal to the SI obtained on Scheme A
→ 111P/100 = 3PR/100 + 3P/4
Solving → R = 12%
Ques 3: A, B, C start a business. The ratio of capitals of A:B is 8:5 and that of B:C is 10:7. If the profit is 1386 at the end of the year, what will be A's share?
This is the ratios question. It will be solved simply.
A:B = 8:5 = 16:10
B:C = 10:7
→ A:B:C = 16:10:7
Now, A's share in profit of 1386 will be = 16*1386/[16+10+7] = 16*1386/33 = 672Rs
Ques 4: The average weight of 32 students is 52.5 Kg. If 8 more students join them, the average becomes 56. Find the average weight of 8 students?
This is average question. Also solved simply.
Average weight of 32 students = 52.5 → total weight of 32 students = 32*52.5 = 1680
8 more students joined → total students now = 40
Their average weight = 56 → total weight of 40 students = 40*56 = 2240
Weight of 8 students will be = 2240-1680 = 560
→ Average weight = 560/8 = 70Kg
Ques 5: The cost of 12 tables and 18 chairs is equal to 3145.What will be the cost of 30 tables and 45 chairs?
12T + 18C = 3145
Now, 30T = 12T*2.5
45C = 18C*2.5
Simply, the cost of 30T + 45C will be 2.5*3145 = 7862.5
(If by multiplying, the value comes unequal, then the cost cannot be determined. For example, we cannot find the cost of 30T and 50C)
Ques 6: Total 471 sweets were to be distributed among students of 5th, 6th and 7th class. Each student got 4 sweets and 87 sweets were left. Find the number of students in 5th class if each class has equal number of students?
Let each class has X students. → 3 classes have 3X students
Now, 3X students got 4 candies each and 87 were left
→ 3X*4 + 87 = 471
→ 12X = 384 → X = 32 Ans
Ques 7: The base and height of a triangle is equal to the length and breadth of a rectangle having perimeter 90 cm. What will be the area of triangle if the difference between length and breadth of the rectangle is 7 cm?
Be simple.
Perimeter of rectangle = 2(L + B) = 90 L + B = 45
And L - B = 7
→ 2L = 52 → L = 26 and B = 19
Now, these were base and height of triangle
Area of triangle = 1/2*B*H = 1/2*26*19 = 247
Ques 8: The ratio between ages of Sarita and Kavita is 11:5. After 6 years, the ratio between the ages of Kavita and Sunita will be 13:10. If age of Sunita was 10 years before 4 years, what is the present age of Sarita?
This is the question of ratios on ages. Solve simply.
Age of Sunita was 10 years before 4 years. → Age now = 14 years (and age after 6 years = 20 years)
Now, Sarita :Kavita = 11:5, let 11X and 5X be their ages
Now, after 6 years, Kavita age = 5X+6, and Sarita = 20 years
→ [5X+6]/20 = 13/10 → 5X+6 = 26 → X = 4
Hence, Present age of Sarita = 11X = 44 years
Ques 9: A person travelled 12.8km in boat 24 minutes in downstream and 14.4km in 36 minutes in upstream. Find the speed of stream?
This question uses this concept
Speed of boat (B) + speed of stream (S) = Speed downstream (D)
Speed of boat (B) - speed of stream (S) = Speed upstream (U)
Hence, B = [D+U]/2 and S = D-B = [D-U]/2
Now, downstream, 12.8km in 24 minutes → 32km in 1 hour
Upstream, 14.4km in 36 minutes → 24km in 1 hour
D and U are calculated → Speed of stream = [32-24]/2 = 4kmph
Ques 10: A pipe can fill a tank in 12 hours, B can empty it in 24 hours. How much more time require to fill the tank if it already filled 1/3rd if both pipes are opened simultaneously?
This is time and work question. Let’s solve it simply.
Let A fills 1 unit of tank in 1 hour → fills it in 12 hour → 12 units of water in tank is there.
Now, B empties it in 24 hour → his speed = 12/24 = 0.5 units per hour
Now, net speed = 1 - 0.5 = 0.5 units per hour
Tank is 1/3 filled → 2/3 empty
2/3 of 12 = 8 units
8 units will be filled with speed of 0.5 units per hour in = 8/0.5 = 16 hours
Ques 11: A is thrice fast as B, together they can finish a work in X days, in how many days B finish the same work alone?
Again, the time and work problem.
Let B does 1 unit per day → A does 3 units per day
→ Both does 4 units per day
Together, they finished the work (with 4 units per day) in 10 days 10*4 = 40 units of work
B finish the work (40 units) alone (1 unit per day) = 40/1 = 40 days
Ques 12: A man sold an item for 3500/- and loss occurred, same article will be sold at 4500/- he would gains same profit as of loss percentage. What was the CP..? (I just make it all these figures just for info, figures are in decimals in exam)
Let CP be X Rs
Profit be Y% = Loss
Now, 4500 = [100+Y]*X/100
And 3500 = [100-Y]*X/100
Solving → Y = 12.5%
And, CP = 4500*100/112.5 = 4000Rs
(This question can also be solved using common sense. on selling at 3500, he had loss and on selling on 4500 he had profit equal to loss → CP = [4500+3500]/2 = 4000Rs)
Ques 13: A man mixed a 25/- Rs rice with 18.50/- Rs rice in the ratio 3:2. Then at what price he should sold the total rice to get a profit of 20%?
Rice mixed in ratio of 3:2
That means per 5 kg, 3 kg is 25rs rice and 2kg is 18.5rs rice.
Now, in 5 kg, he had invested Rs = 25*3 + 18.5*2 = 75+37 = 112Rs
For profit of 20%, he should sold this 5kg rice in Rs = [120/100]*112 = 134.4Rs
134.4 Rs for 5kg → 134.4/5 = 26.88Rs per kg
Ques 14: A shopkeeper marks his article 50% above the cost price and then offers a discount of 28% on marked price. Find his overall profit percentage?
Let cost price be 100X Rs.
→ MP = 1.5*100X = 150X Rs
Discount = 28% of 150X = 42X
→ SP = 150X - 42X = 108X
→ Profit = 8X on 100X Profit % = 8%
Ques 15: A certain amount of money amounts to 12040 after 2 years at compound interest, and amounts to 13244 after 3 years, what is rate of interest?
This question can be solved quickly using any of these approach.
Approach 1:
Equation (1) → P*[1 + R/100]^3 = 13244
Equation (2) → P*[1 + R/100]^2 = 12040
Dividing equation (1) by (2), we get,
1 + R/100 = 13244/12040 = 1 + 1204/12040 → R/100 = 1/10 → R = 10%
Approach 2:
12040Rs became 13244 Rs in 1 year. That means 1204Rs interest on 12040Rs in 1 year. This can be solved using SI formula
Interest = PRT/100, where P=12040, R=unknown, T=1year
→ 1204 = 12040*R*1/100 → R = 10%
Ques 16: A grocer sells mixture of two verities of Rice at Rs 56.25 per kg thereby having 25% profit. He mixes 72 kg of brand A verity costing 49 per kg with 40 kg of another verity of brand B. what is price per kg of brand B rice?
Let’s solve it using basic concepts.
56.25 Rs per Kg is SP then CP = 56.25/1.25 = 45Rs per Kg
Now, 72 Kg of brand A cost Rs 49 → 72*49 = 3528 Rs
40 Kg of brand B cost Rs X → 40X Rs
Mixture 72+40 = 112 Kg of rice cost Rs 45 per Kg → 5040 Rs
Now, 3528 + 40X = 5040 → 40X = 1512 → X = 37.8 Rs = Ans
Ques 1: The difference between 75% and 4/7 of a number is 125. Find the number.
This question is solved simply.
Let the number be X
Then 75% of X becomes 3X/4
Now, 3X/4 - 4X/7 = 125
Cross multiply → 5X/28 = 125 → X = 25*28 = 700
Ques 2: An amount is invested in scheme A for 6 years at simple interest at the rate of interest was 18.5%. If 3/4th of same amount is invested for 4 years in scheme B on simple interest, it becomes equal to the interest obtained in scheme A. What is the rate of interest in scheme B?
This question is based upon the basic concepts of simple interest.
SI = PRT/100
Now, let the amount invested (Principal) in Scheme A be P Rs.
According to question, SI in Scheme A = P*18.5*6/100 = 111P/100
Now, in second case, the amount invested in Scheme B is 3/4 of P = 3P/4
Now, SI on 3P/4 for 4 years will be (taking rate of interest as R%) = ([3P/4]*R*4)/100 = 3PR/100
This is SI, then final amount after we add SI to 3P/4 = 3PR/100 + 3P/4
This amount is equal to the SI obtained on Scheme A
→ 111P/100 = 3PR/100 + 3P/4
Solving → R = 12%
Ques 3: A, B, C start a business. The ratio of capitals of A:B is 8:5 and that of B:C is 10:7. If the profit is 1386 at the end of the year, what will be A's share?
This is the ratios question. It will be solved simply.
A:B = 8:5 = 16:10
B:C = 10:7
→ A:B:C = 16:10:7
Now, A's share in profit of 1386 will be = 16*1386/[16+10+7] = 16*1386/33 = 672Rs
Ques 4: The average weight of 32 students is 52.5 Kg. If 8 more students join them, the average becomes 56. Find the average weight of 8 students?
This is average question. Also solved simply.
Average weight of 32 students = 52.5 → total weight of 32 students = 32*52.5 = 1680
8 more students joined → total students now = 40
Their average weight = 56 → total weight of 40 students = 40*56 = 2240
Weight of 8 students will be = 2240-1680 = 560
→ Average weight = 560/8 = 70Kg
Ques 5: The cost of 12 tables and 18 chairs is equal to 3145.What will be the cost of 30 tables and 45 chairs?
12T + 18C = 3145
Now, 30T = 12T*2.5
45C = 18C*2.5
Simply, the cost of 30T + 45C will be 2.5*3145 = 7862.5
(If by multiplying, the value comes unequal, then the cost cannot be determined. For example, we cannot find the cost of 30T and 50C)
Ques 6: Total 471 sweets were to be distributed among students of 5th, 6th and 7th class. Each student got 4 sweets and 87 sweets were left. Find the number of students in 5th class if each class has equal number of students?
Let each class has X students. → 3 classes have 3X students
Now, 3X students got 4 candies each and 87 were left
→ 3X*4 + 87 = 471
→ 12X = 384 → X = 32 Ans
Ques 7: The base and height of a triangle is equal to the length and breadth of a rectangle having perimeter 90 cm. What will be the area of triangle if the difference between length and breadth of the rectangle is 7 cm?
Be simple.
Perimeter of rectangle = 2(L + B) = 90 L + B = 45
And L - B = 7
→ 2L = 52 → L = 26 and B = 19
Now, these were base and height of triangle
Area of triangle = 1/2*B*H = 1/2*26*19 = 247
Ques 8: The ratio between ages of Sarita and Kavita is 11:5. After 6 years, the ratio between the ages of Kavita and Sunita will be 13:10. If age of Sunita was 10 years before 4 years, what is the present age of Sarita?
This is the question of ratios on ages. Solve simply.
Age of Sunita was 10 years before 4 years. → Age now = 14 years (and age after 6 years = 20 years)
Now, Sarita :Kavita = 11:5, let 11X and 5X be their ages
Now, after 6 years, Kavita age = 5X+6, and Sarita = 20 years
→ [5X+6]/20 = 13/10 → 5X+6 = 26 → X = 4
Hence, Present age of Sarita = 11X = 44 years
Ques 9: A person travelled 12.8km in boat 24 minutes in downstream and 14.4km in 36 minutes in upstream. Find the speed of stream?
This question uses this concept
Speed of boat (B) + speed of stream (S) = Speed downstream (D)
Speed of boat (B) - speed of stream (S) = Speed upstream (U)
Hence, B = [D+U]/2 and S = D-B = [D-U]/2
Now, downstream, 12.8km in 24 minutes → 32km in 1 hour
Upstream, 14.4km in 36 minutes → 24km in 1 hour
D and U are calculated → Speed of stream = [32-24]/2 = 4kmph
Ques 10: A pipe can fill a tank in 12 hours, B can empty it in 24 hours. How much more time require to fill the tank if it already filled 1/3rd if both pipes are opened simultaneously?
This is time and work question. Let’s solve it simply.
Let A fills 1 unit of tank in 1 hour → fills it in 12 hour → 12 units of water in tank is there.
Now, B empties it in 24 hour → his speed = 12/24 = 0.5 units per hour
Now, net speed = 1 - 0.5 = 0.5 units per hour
Tank is 1/3 filled → 2/3 empty
2/3 of 12 = 8 units
8 units will be filled with speed of 0.5 units per hour in = 8/0.5 = 16 hours
Ques 11: A is thrice fast as B, together they can finish a work in X days, in how many days B finish the same work alone?
Again, the time and work problem.
Let B does 1 unit per day → A does 3 units per day
→ Both does 4 units per day
Together, they finished the work (with 4 units per day) in 10 days 10*4 = 40 units of work
B finish the work (40 units) alone (1 unit per day) = 40/1 = 40 days
Ques 12: A man sold an item for 3500/- and loss occurred, same article will be sold at 4500/- he would gains same profit as of loss percentage. What was the CP..? (I just make it all these figures just for info, figures are in decimals in exam)
Let CP be X Rs
Profit be Y% = Loss
Now, 4500 = [100+Y]*X/100
And 3500 = [100-Y]*X/100
Solving → Y = 12.5%
And, CP = 4500*100/112.5 = 4000Rs
(This question can also be solved using common sense. on selling at 3500, he had loss and on selling on 4500 he had profit equal to loss → CP = [4500+3500]/2 = 4000Rs)
Ques 13: A man mixed a 25/- Rs rice with 18.50/- Rs rice in the ratio 3:2. Then at what price he should sold the total rice to get a profit of 20%?
Rice mixed in ratio of 3:2
That means per 5 kg, 3 kg is 25rs rice and 2kg is 18.5rs rice.
Now, in 5 kg, he had invested Rs = 25*3 + 18.5*2 = 75+37 = 112Rs
For profit of 20%, he should sold this 5kg rice in Rs = [120/100]*112 = 134.4Rs
134.4 Rs for 5kg → 134.4/5 = 26.88Rs per kg
Ques 14: A shopkeeper marks his article 50% above the cost price and then offers a discount of 28% on marked price. Find his overall profit percentage?
Let cost price be 100X Rs.
→ MP = 1.5*100X = 150X Rs
Discount = 28% of 150X = 42X
→ SP = 150X - 42X = 108X
→ Profit = 8X on 100X Profit % = 8%
Ques 15: A certain amount of money amounts to 12040 after 2 years at compound interest, and amounts to 13244 after 3 years, what is rate of interest?
This question can be solved quickly using any of these approach.
Approach 1:
Equation (1) → P*[1 + R/100]^3 = 13244
Equation (2) → P*[1 + R/100]^2 = 12040
Dividing equation (1) by (2), we get,
1 + R/100 = 13244/12040 = 1 + 1204/12040 → R/100 = 1/10 → R = 10%
Approach 2:
12040Rs became 13244 Rs in 1 year. That means 1204Rs interest on 12040Rs in 1 year. This can be solved using SI formula
Interest = PRT/100, where P=12040, R=unknown, T=1year
→ 1204 = 12040*R*1/100 → R = 10%
Ques 16: A grocer sells mixture of two verities of Rice at Rs 56.25 per kg thereby having 25% profit. He mixes 72 kg of brand A verity costing 49 per kg with 40 kg of another verity of brand B. what is price per kg of brand B rice?
Let’s solve it using basic concepts.
56.25 Rs per Kg is SP then CP = 56.25/1.25 = 45Rs per Kg
Now, 72 Kg of brand A cost Rs 49 → 72*49 = 3528 Rs
40 Kg of brand B cost Rs X → 40X Rs
Mixture 72+40 = 112 Kg of rice cost Rs 45 per Kg → 5040 Rs
Now, 3528 + 40X = 5040 → 40X = 1512 → X = 37.8 Rs = Ans
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