Dear Aspirants, Quantitative Aptitude plays a crucial role in Banking and all other competitive exams. To enrich your preparation, here we have provided New Pattern Aptitude Questions for IBPS Clerk Mains. Candidates those who are going to appear in IBPS Clerk Mains can practice these questions daily and make your preparation effective.
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Directions (1-5): Study the following pie-chart and table carefully and answer the questions:
The Table below shows the discount % given by different shops A1, A2, A3 and the following line graph shows the profit % gained on the given articles by the given stores. It is given that Market Price of any article at every shop is same.
1)
Quantity I: Find the marked price of an article in Shop A1, if the average cost price of an article 4 and 5 is Rs. 5550.
Quantity II: Find the marked price of an article in Shop A2, if the cost price of an article 2 Rs. 820 more than the cost price of an article 3
Quantity III: Find the marked price of an article in Shop A3, if the sum cost price of an article 1 and 4 is Rs. 19375
Which of the following should be placed in the blank spaces of the expression “Quantity I __ Quantity II ___ Quantity III” from left to right with respect to the above statements?
a) <, <
b) >, <
c) >, >
d) =, >
e) <, =
2) Find the CP of Article 4 if the average SP of this article in all the shops is Rs. 7710?
a) Rs 18504
b) Rs 19275
c) Rs 37014
d) Rs 27028
e) Rs 17765
3) What is the approximate difference between the CP of Article 3 of Shop A2 and Article 2 of Shop A3 if the MP is Rs. 2450?
a) Rs 235
b) Rs 255
c) Rs 217
d) Rs 139
e) Rs 282
4) If the CP of Article 2 of Shop A2 is greater than CP of Article 1 of Shop A3 by 350. What is the difference between the SP of Article 1 of Shop A2 and SP of Article 2 of A1?
a) Rs 4416
b) Rs 3818
c) Rs 4254
d) Rs 4602
e) Rs 3625
5) If Article 6 is introduced in Shop A1 with SP of Rs 4560 and if it equals the CP of Article 2 of same shop. What will be the approximate discount % on it?
a) 15 %
b) 20 %
c) 25 %
d) 37 %
e) 29%
Directions (6 – 10): What should come at the place of question mark in the following questions?
6) 1/7 x 1/5 x 4/9 x 3150 + 40% of 7800 =? – 540
a) 3700
b) 5600
c) 4400
d) 3200
e) None of these
7) 1/19 x 7638 + 1/26 x 2080 =? x 241
a) 3
b) 6
c) 4
d) 2
e) None of these
8) 343.55 + 533.43 – 4444.35 + 3353.54 =? – 333.33
a) 112.5
b) 114.5
c) 119.5
d) 110.5
e) None of these
9) √5776/4 – √4225 + √9604 x 5 =?
a) 444
b) 626
c) 740
d) 582
e) None of these
10) 80% of 24000 – 16% of 7200 =? – 15% of 2000
a) 12360
b) 18348
c) 20650
d) 16408
e) None of these
Click Below for Video Solutions for these Questions
Answers :
Direction (1-5) :
1) Answer: a)
From Quantity I: Find the marked price of an article in Shop A1, if the average cost price of an article 4 and 5 is Rs. 5550.
Let us take the marked price of an article be M in shop A1,
SP of an article 4 = 90M/100
CP of an article 4 = 90M/100 *(100/120)
SP of an article 5 = 80M/100
CP of an article 5 = 80M/100 *(100/140)
Total CP of an article 4 and 5 in shop A1 = 5550 * 2 = Rs.11100
[90M/100*(100/120)] + [80M/100*(100/140)] = 11100(90M/120 + 80M/140) = 11100
(630M+480M)/840 = 11100
1110M/840 = 11100
M = Rs. 8400
From Quantity II: Find the marked price of an article in Shop A2, if the cost price of an article 2 Rs. 820 more than the cost price of an article 3
Let us take the marked price of an article be M in shop A2
SP of an article 2 = 85M/100
CP of an article 2 = 85M/100 *(100/120)
SP of an article 3 = 80M/100
CP of an article 3 = 80M/100 *(100/125)
[85M/100 *(100/120)] – [80M/100 *(100/125)] = 82085M/120 – 80M/125 = 820
2125M – 1920M = 820 *3000
205M = 820*3000
M = Rs.12000
From Quantity III: Find the marked price of an article in Shop A3, if the sum cost price of an article 1 and 4 is Rs. 19375
Let us take the marked price of an article be M in shop A3
SP of an article 1 = 80M/100
CP of an article 1= 80M/100 *(100/115)
SP of an article 4 = 85M/100
CP of an article 4 = 85M/100 *(100/120)
Total CP of an article 1 and 4 in shop A3 = 19375
[80M/100*(100/115)] + [85M/100*(100/120)] = 1937580M/115 + 85M/120 = 19375
1920M + 1955M = 19375 * 2760
3875M = 19375 * 2760
M = Rs.13800
Hence, Quantity I < Quantity II < Quantity III
2) Answer: b)
Let the MP of Article 4 be M,
SP of Shop A1 of Article 4 = 90M/100
SP of Shop A2 of Article 4 = 82M/100
SP of Shop A3 of Article 4 = 85M/100
Average of SP of Shop 1, 2 and 3 of Article
(i.e) ((90M/100) + (82M/100) + (85M/100))/3 = 7710
Solving the above equation, M = 9000
Total SP of Article 4= (90*9000/100) + (82*9000/100) + (85*9000/100)
= 257 * 9000/100
= 23130
Thus required CP = 23130*(100/120) = Rs. 19275
3) Answer: d)
SP of Article 3 of shop A2 = (80*2450)/100
= Rs 1960
CP of Article 3 of Shop A2 = (1960 * 100)/125 = Rs. 1568
Again,
SP of Article 2 of Shop A3 = (70 * 2450)/100
= Rs 1715
CP of Article 2 of Shop A3 = (1715 * 100)/120 = Rs 1429.16
Difference between the CP of Article 3 of Shop A2 and Article 2 of Shop A2
= > 1568 – 1429.16 = Rs. 139
4) Answer: a)
Let the MP be M,
CP of Article 2 of Shop A2 = (85M/100)*(100/120)
= 17M/24
CP of Article 1 of Shop A3 = (80M/100)* (100/115)
= 16M/23
As given in the Question,
17M/24 – 16M/23 = 350
M = 27600
Now, SP of Article 1 of Shop A2 = (92*27600)/100 = 25392
Again SP of Article 2 of Shop A1= (76*27600)/100
= 20976
Difference of SP of Article 1 of Shop A2 and SP of Article 2 of Shop A1
= 25392 – 20976 = Rs 4416
5) Answer: d)
Let the MP be M and discount be x %,
CP of Article 2 of Shop A1 = (76M/100) * (100/120)
As per question,
(76M/100) *(100/120) = 4560
So, M = 7200
Again,
SP of Article 6 of Shop A1 = (100 – x) * 7200/100
(ie.,) (100 – x) * 7200/100 = 4560
100 – x = 4560*(100/7200)
100 – x = 63.33
x = 36.67 % = 37 %
Direction (6-10) :
6) Answer: a)
1/7 x 1/5 x 4/9 x 3150 + 40% of 7800 = ? – 540
= > 40 + (40/100) x 7800 + 540 = ?
= > 40 + 3120 + 540 = ?
= > 3700 = ?
7) Answer: d)
(1/19) x 7638 + (1/26) x 2080 =? x 241
= > 402 + 80 =? x 241
= > 482/241 =?
= > 2 =?
8) Answer: c)
343.55 + 533.43 – 4444.35 + 3353.54 =? – 333.33
= > – 213.83 =? – 333.33
= > 333.33 – 213.83 =?
= > 119.5 =?
9) Answer: a)
√5776/4 – √4225 + √9604 x 5 =?
= > 76/4 – 65 + 98 x 5 =?
= > 19 – 65 + 490 =?
= > 444 =?
10) Answer: b)
80% of 24000 – 16% of 7200 =? – 15% of 2000
= > (80/100)*24000 – (16/100)*7200 =? – (15/100)*2000
= > 19200 – 1152 =? – 300
= > 19200 – 1152 + 300 =?
= > 18348 =?
Daily Practice Test Schedule | Good Luck
| Topic | Daily Publishing Time |
| Daily News Papers & Editorials | 8.00 AM |
| Current Affairs Quiz | 9.00 AM |
| Current Affairs Quiz (Hindi) | 9.30 AM |
| IBPS SO/NIACL AO Prelims – Reasoning | 10.00 AM |
| IBPS SO/NIACL AO Prelims – Reasoning (Hindi) | 10.30 AM |
| IBPS SO/NIACL AO Prelims – Quantitative Aptitude | 11.00 AM |
| IBPS SO/NIACL AO Prelims – Quantitative Aptitude (Hindi) | 11.30 AM |
| Vocabulary (Based on The Hindu) | 12.00 PM |
| IBPS SO/NIACL AO Prelims – English Language | 1.00 PM |
| SSC Practice Questions (Reasoning/Quantitative aptitude) | 2.00 PM |
| IBPS Clerk – GK Questions | 3.00 PM |
| SSC Practice Questions (English/General Knowledge) | 4.00 PM |
| Daily Current Affairs Updates | 5.00 PM |
| SBI PO/IBPS Clerk Mains – Reasoning | 6.00 PM |
| SBI PO/IBPS Clerk Mains – Quantitative Aptitude | 7.00 PM |
| SBI PO/IBPS Clerk Mains – English Language | 8.00 PM |
