Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From DATA SUFFICIENCY topic, which is common for all the IBPS,SBI exam RBI,,SSC and other competitive exams. We have included Some questions that are repeatedly asked in exams !!!
Q(1 – 10) Each of the questions below consists of a question and three statements given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the three statements and give answer: –
- Reena and Meera bought and sold some sarees. What is the cost price of each saree?
I. The selling price of 29 sarees is 7395.
II. They had sold N sarees, where the difference between N and the number obtained by interchanging its digits is 9.
III. Even though they spent 150 on packaging, they made a total profit of 20%.
IV. The number of sarees sold is the first two digit number which is preceded and followed by prime numbers.
a. I, II, III
b. All the four
C. I, III and II or IV
d. I , III and IV
e. None of the aboveAnswer – d. I , III and IV
I –> sp of one saree = 255
III –> sp = ((N*cp )+ 150)/N * 120%
IV –> 2,3,4,5,6,7,8,9,10,11,12,13. 12 is the first two digit number followed and preceded by prime numbers.
Therefore , CP = 200.
- What are the ages of Lia and Mia?
I. Three years ago, Lia was 5 years younger to Mangesh and Mia was 7 years older to Mangesh.
II. Ratio of present ages of Lia and Mia is 15:19
III. Sum of the ages of Mangesh and Lia is 95
a. I & II
b. I &III
c. I, II & III
d. I & II or III
e. None of theseAnswer – d. I & II or III
I –> Mangesh – Lia = 5
Mia – Mangesh = 7
=> Mia –Lia = 12
II –> Lia : Mia = 15:19
4 ratio =12
1 ratio = 3
Therefore, Lia = 45 & Mia = 57
III –> Mangesh + Lia = 95
=> Lia’s age = (95-5)/2 = 45
Therefore , Mia’s age = 45=12 = 57
- There were two traders, Matthew and Perry. They bought some chairs and tables from Stephen. Even though Stephen was getting a profit of 200 Rs/ chair and 370 Rs/ table, he charged Matthew and Perry an extra 50Rs per piece as handling charges. For what price did Matthew and Perry sell the chair, if they earned a total profit of 23.76%?
I. Total amount spent on the chairs and tables were 1,59,000 and 3,21,500 Rs respectively.
II. There were equal number of chairs and tables and they sold only half of them.
III. The selling price of 5 tables and 2 chairs was 92,950.
a. All the three
b. I & II
c. I & III
d. II & III
e. None of the aboveAnswer – e. None of the above
None of the given information can be connected to the SP or CP of the chairs and tables.
- What are the marks obtained by Sreya and Kia(assuming Sreya scored less than Kia)?
I. The sum of their marks form 70% of the total score.
II. Ratio of marks of Sreya and Kia is 29:41
III. Ratio of the sum of their marks to difference of their marks is 35:6
IV. The square of six fifth of three fifth of half of Sreya’s mark is 68121
a. I & IV
b. I,II & IV
c. I,II & III
d. II or III, I & IV
e. II or III, & IVAnswer – e. II or II & IV
IV –> Sreya’s mark = sqrt(68121) * 2* 5/3 *5/6 = 725.
II –> S: K = 29:41
29 ratio = 725 => 1 ratio = 25
Therefore , Kia’s mark = 41*25 = 1025.
III —>( S+K) : (K-S) = 35:6 => S:K = 29:41
- What will be the population of town Z in the year 2052?
I. Present population of the town is 18,85,200
II. Population of the town 23 years back was 2,97,021
III. Rate of increase of population of town Z is 13.425%
IV. The ratio of the men to women ratio in 2052 will be 17:23
a. All are needed
b. Only I & III
c. None of the statements is sufficient
d. Only I & II
e. I, II & IIIAnswer – c. None of the statements is sufficient
Question asks for the population of town Z in 2052. But none of the statements say which is the current year. Therefore, the population in 2052 cannot be determined with the above details.
- A and B together completed a work in 6 days. They got Rs. 1350 as remuneration. What would be the ratio in which the 1350 Rs would be split if C worked along with A and B?
I. Ratio of no: of days taken by A & C is 5:9
II. Ratio of efficiency of A & B = 3:2
III. C is the least efficient of all
a. I only
b. II only
c. I & II
d. I, II & III
e. None of the aboveAnswer – c. I & II
Days 5x 9x
efficiency 3y 2y
Total work / (3y+2y) = 6 => 30y = TW
5 x —->3y
=> Efficiency of C = 5/3 y
Therefore ratio = 3:2:5/3 = 9:6:5
- Gina got 50 marks in English, but failed in the examination. What was the pass marks that year?
I. Malu who got 80 passed the exam.
II. Total marks of the examination was 190
III. If Malu could give Gina the extra marks( than what she needed to pass the exam) she had, Gina would have passed with 10 extra marks.
a. I & II
b. I, II & III
c. I & III
d. II & III
e. NoneAnswer – C. I & III
Pass mark = 50+ (80-10-50)/2 = 60.
- What is the amount invested by Sanchi in the business?
I. Ratio of amounts invested by Sachi and Ada = 9:11
II. Total profit obtained = 1418520 Rs
III. Amount invested by Ada constitutes 55% of the total investment
IV. Amount invested by Ada is double the square of triple the square root of 3025
a. I, II & III
b. I & III or IV
c. III & IV
d. I or II & IV
e. I & IV or III & IVAnswer – E. I or III & IV
I — -> S :A = 9:11
III —> A:S = 11:9
VI —–> Amount invested by Ada =(((sqrt(3025))*3)^2)*2 = 54,450 Rs
Amount invested by Sachi = 54450* 9/11 = 44550 Rs.
- What are the values of x and y?
I. 21x + 33y + 9 = 180
II. x+y+249 = 256
III. (x-y)*87 = 261
a. I & II or III
b. II & I or III
c. I, III & III
d. I & II or I & III or II & III
e. None of the aboveAnswer – d. I & II or I & III or II & III
By solving any two of the given linear equations, you’ll get x = 5 & y= 2
- What is the area of the plot which is in the shape of a regular polygon?
I. Perimeter of the plot is 1036m
II. Radius of the largest circle that can be inscribed in the plot is 129.5m
III. Diameter of the largest circle that can be inscribed in the plot is equal to the length one side of the plot.
a. I, II & III
b. I or II, & III
c. I & II
d. None of the aboveAnswer – b. I or II, & III
From III, it is clear that the plot is square, as the side of the regular polygon shaped plot is equal to the diameter of the biggest circle that can be inscribed in it.
From I, 4a = 1036m
=> a = 259m
=> area = 259*259 = 67081sq.m
From II, radius of circle = 129.5m
Therefore, diameter = a = 129.5*2 = 259m
=> area = 67081 sq.m