Algebra - Simple Equations - Previous Year CAT/MBA Questions
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Aditya has a total of 18 red and blue marbles in two bags (each bag has marbles of both colors). A marble is randomly drawn from the first bag followed by another randomly drawn from the second bag, the probability of both being red is 5/16. What is the probability of both marbles being blue?
- (a)
- (b)
- (c)
- (d)
4 16
- (e)
None of the above
Answer: Option C
Text Explanation :
(X1 + Y1) + (X2 + Y2) = 18 … (i)
Selecting a marble from the first bag and then from the second bag can be done in (X1 + Y1) × (X2 + Y2) ways.
Selecting a red marble from the first bag and then a red marble from the second bag can be done in (X1) × (X2) ways.
∴ Probability of selecting red marbles from both the bags = (X1) × (X2)/ (X1 + Y1) × (X2 + Y2) = 5/16
Let (X1 + Y1) × (X2 + Y2) = 16a … (ii)
∴ (X1) × (X2) = 5a … (iii)
Considering (i) and (ii), a = 2 or 5
Case I: a = 2
(X1 + Y1) × (X2 + Y2) = 32 … (iv)
∴ From (i) and (iv),
(X1 + Y1) = 2 and (X2 + Y2) = 16
∴ X1 = Y1 = 1 ⇒ X2 = 10 (∵ (X1) × (X2) = 10)
∴ Y2 = 6
Probability of both marbles being blue = (1 × 6)/32
= 3/16
Case II: a = 5
(X1 + Y1) × (X2 + Y2) = 80 … (v)
∴ From (i) and (v),
(X1 + Y1) = 8 and (X2 + Y2) = 10
∴ X1 = X2 = 5 (∵ (X1) × (X2) = 25)
Y1 = 3 and Y2 = 5
∴ Y2 = 6
Probability of both marbles being blue = (5 × 3)/80
= 3/16
Hence, option (c).
Workspace:
Consider a rectangle ABCD of area 90 units. The points P and Q trisect AB, and R bisects CD. The diagonal AC intersects the line segments PR and QR at M and N respectively. What is the area of the quadrilateral PQMN?
- (a)
> 9.5 and ≤ 10
- (b)
> 10 and ≤ 10.5
- (c)
> 10.5 and ≤ 11
- (d)
> 11 and ≤ 11.5
- (e)
> 11.5
Answer: Option D
Text Explanation :
Let the breadth be 3x and the breadth be y.
3xy = 90 ⇒ xy = 30
V is midpoint of WR. PW || EV ⇒ EV = PW/2
Similarly, FV = WQ/2
∴ EF = PQ/2 = x/2
∆MPA ∼ ∆MEV
Height of ∆MPA with respect to AP: Height (h1)of ∆MEV with respect to EV = AP : EV = x : x/4 = 4 : 1
Let height of ∆MPA = 4k and height (h1) of ∆MEV = k
∴ 4k + k = 5k = y/2
∴ k = y/10
∴ Height (h1) of ∆MEV = y/10
Similarly, ∆VFN ∼ ∆CRN
Height (h2) of ∆VFN with respect to VF : Height of ∆CRN with respect to CR = VF : CR = x/4 : 3x/2 = 1 : 6
Let height (h2) of ∆VFN = m and height of ∆CRN = 6m
∴ m + 6m = 7m = y/2
∴ m = y/14
∴ Height (h2) of ∆VFN = y/14
A(□PQMN) = A(□PQEF) – A(∆MEV) + A(∆VFN)
= 11.25 – 0.375 + 0.27 ≈ 11.145
Hence, option (d).
Workspace:
Two numbers, 297B and 792B, belong to base B number system. If the first number is a factor of the second number then the value of B is:
- (a)
11
- (b)
12
- (c)
15
- (d)
17
- (e)
19
Answer: Option E
Text Explanation :
Solving by options,
By option(1), we get,
297B = 112 × 2 + 11 × 9 + 7 = 348
792B = 112 × 7 + 11 × 9 + 2 = 948
Since 348, is not a factor of 948, option(1) is eliminated.
By option(2), we get,
297B = 122 × 2 + 12 × 9 + 7 = 403
792B = 122 × 7 + 12 × 9 + 2 = 1118
Since 403, is not a factor of 1118, option(2) is eliminated.
By option(3), we get,
297B = 152 × 2 + 15 × 9 + 7 = 592
792B = 152 × 7 + 15 × 9 + 2 = 1712
Since 592, is not a factor of 1712, option(3) is eliminated.
By option(4), we get,
297B = 172 × 2 + 17 × 9 + 7 = 738
792B = 172 × 7 + 17 × 9 + 2 = 2178
Since 738, is not a factor of 2178, option(4) is eliminated.
By option(5), we get,
297B = 192 × 2 + 19 × 9 + 7 = 900
792B = 192 × 7 + 19 × 9 + 2 = 2700
Since 900, is a factor of 2700,
Hence, option (e).
Workspace:
A teacher noticed a strange distribution of marks in the exam. There were only three distinct scores: 6, 8 and 20. The mode of the distribution was 8. The sum of the scores of all the students was 504. The number of students in the in most populated category was equal to the sum of the number of students with lowest score and twice the number of students with the highest score. The total number of students in the class was:
- (a)
50
- (b)
51
- (c)
53
- (d)
56
- (e)
57
Answer: Option E
Text Explanation :
Let the number of students scoring 6, 8 and 20 be x, y and z respectively.
So, 6x + 8y + 20z = 504
x + 2z = y
or, 14y + 8z = 504
or, 7y + 4z = 252
By hit and trial we get y = 32 and z = 7
Therefore, x = 18
Therefore, total number of students = 32 + 7 + 18 = 57
Hence, option (e).
Workspace:
Read the following instruction carefully and answer the question that follows:
Expression
can also be written as
What would be the remainder if x is divided by 11?
- (a)
2
- (b)
4
- (c)
7
- (d)
9
- (e)
None of the above
Answer: Option D
Text Explanation :
All the terms in x are divisible by 11 except 13!/11
According to Wilson theorem,
Hence, option (d).
Workspace:
A rectangular swimming pool is 48 m long and 20 m wide. The shallow edge of the pool is 1 m deep. For every 2.6 m that one walks up the inclined base of the swimming pool, one gains an elevation of 1 m. What is the volume of water (in cubic meters), in the swimming pool? Assume that the pool is filled up to the brim.
- (a)
528
- (b)
960
- (c)
6790
- (d)
10560
- (e)
12960
Answer: Option D
Text Explanation :
Since the elevation increases by 1 m for every 2.6 m,
Since
The height of the deeper end of the pool is 20 m.
This can be represented as follows,
The total volume of water in the pool,
⇒ 48× 20 × 11
⇒ 10560 m³
Hence, option (d).
Workspace:
The value of the expression:
- (a)
0.01
- (b)
0.1
- (c)
1
- (d)
10
- (e)
100
Answer: Option C
Text Explanation :
Hence, option (c).
Workspace:
Answer the questions based on the trends lines from the following graphs.
Note: Left side of X axis represents countries that are “poor” and right side of X axis represents countries that are “rich”, for each region. GDP is based on purchasing power parity (PPP).
These are World Bank (WB) estimates.
Which of the following could be the correct ascending order of democratic regions for poor?
- (a)
North America, C and E Europe, South America, Middle East, Asia Pacific
- (b)
Scandinavia, Western Europe, Africa, Asia Pacific, Middle East
- (c)
Scandinavia, Western Europe, North America, C and E Europe, Middle East
- (d)
C and E Europe, Africa, South America, Western Europe, Scandinavia
- (e)
Africa, South America, Western Europe, North America, Scandinavia
Answer: Option D
Text Explanation :
Observe that Scandinavia and Western Europe have all their democratic score points as 20. SO, they should be the last two regions in ascending order. However, Scandinavia has no difference between rich and poor. So, it should be the last region and Western Europe should be second last.
Only option 4 satisfies this condition.
Hence, option (d).
Workspace:
Which region has the highest disparity, of democratic participation, between rich and poor?
- (a)
North America
- (b)
C and E Europe
- (c)
Africa
- (d)
South America
- (e)
Western Europe
Answer: Option B
Text Explanation :
The greatest disparity of democratic participation between rich and poor is obtained from the graph with the steepest slope and having some point as close as possible to full democracy and another point as close as possible to full authoritarianism. This condition is satisfied for C and E Europe where the slope of the line is the steepest and the difference between the maximum and minimum points is around 18.
Hence, option (b).
Workspace:
The maximum GDP of African region is higher than the maximum GDP of South American region by factor of:
- (a)
10
- (b)
100
- (c)
2
- (d)
4
- (e)
None of these
Answer: Option E
Text Explanation :
The maximum GDP of Africa and South America has the same value.
Hence, option (e).
Workspace:
Answer the questions based on the given data on the tourism sector in India.
In which of the following years the percentage increase in the number of Indians going abroad was greater than the percentage increase in the number of domestic tourists?
- (a)
2004 and 2005
- (b)
2005 and 2006
- (c)
2005 and 2007
- (d)
2006 and 2008
- (e)
2004, 2005 and 2006
Answer: Option C
Text Explanation :
Observe that the required percentage is to be found for all years from 2004 to 2008.
The percentage increase in the number of Indians going abroad and the percentage increase in the number of domestic tourists for each year from 2004 to 2008 is as shown below:
Thus, the percentage increase in the number of Indians going abroad is greater than the percentage increase in the number of domestic tourists in 2005, 2007 and 2008.
Hence, option (c).
Workspace:
In which of the following years was the rupee cheapest with respect to the dollar?
- (a)
2001
- (b)
2002
- (c)
2007
- (d)
2010
- (e)
2011
Answer: Option B
Text Explanation :
The rupee is cheapest with respect to the dollar when the numerical value of rupees corresponding to 1 dollar is the highest
i.e. If 1 US $ = X INR, then the rupee is cheapest when X takes its largest numerical value.
The value of the rupee with respect to the dollar for the given years is:
Thus, the rupee is cheapest in 2002.
Hence, option (b).
Workspace:
Let ‘R’ be the ratio of Foreign Exchange Earnings from Tourism in India (in US $ million) to Foreign Tourist Arrivals in India (in million). Assume that R increases linearly over the years. If we draw a pie chart of R for all the years, the angle subtended by the biggest sector in the pie chart would be approximately:
- (a)
24
- (b)
30
- (c)
36
- (d)
42
- (e)
48
Answer: Option C
Text Explanation :
The value of R for each year is as shown below:
When a pie-chart is made using the values of R, the value for 2011 forms the largest sector. Also, the total value of R is 26415.49.
Observe that the value of R in 2011 is approximately 10% of the total. So, the angle that it subtends at the centre is 10% of 360 i.e. 36 degrees.
Hence, option (c).
Workspace:
Answer the questions based on the following information:
The exhibit given below compares the countries (first column) on different economic indicators (first row), from 2000-2010. A bar represents data for one year and a missing bar indicates missing data. Within an indicator, all countries have same scale.
Which of the following countries, after United States, has the highest spending on military as % of GDP, in the period 2000-2010?
- (a)
Vietnam
- (b)
China
- (c)
India
- (d)
Brazil
- (e)
Thailand
Answer: Option C
Text Explanation :
It is observed from the bar-graphs that India has the second-highest spending on military as percentage of GDP in the period 2000-2010.
Hence, option (c).
Workspace:
Which country (and which year) has witnessed maximum year-to-year decline in “industry as percentage of GDP”? Given that the maximum value of industry as percentage of GDP is 49.7% and the minimum value of industry as percentage of GDP is 20.02%, in the chart above.
- (a)
United States in 2002-3
- (b)
Brazil in 2006-7
- (c)
India in 2009-10
- (d)
Malaysia in 2008-9
- (e)
China in 2008-9
Answer: Option D
Text Explanation :
Among the options given, it can be seen that Malaysia in 2008-09 has witnessed the maximum year-on-year decline in “industry as percentage of GDP”.
Hence, option (d).
Workspace:
Which of the following countries has shown maximum increase in the “services, value added as % of GDP” from year 2000 to year 2010?
- (a)
Brazil
- (b)
India
- (c)
United States
- (d)
Philippines
- (e)
None of the above
Answer: Option B
Text Explanation :
It can be seen that India has shown the maximum increase in the “services, value added as percentage of GDP” from 2000 to 2010”.
Hence, option (b).
Workspace:
The sum of 1 - + - + ... is
- (a)
- (b)
- (c)
- (d)
Answer: Option D
Text Explanation :
1 - + × - × × .....
1 - =
Observe that the sum of further terms will be positive.
Hence, the sum will be greater than 5/6 i.e. 0.833.
Also, if we add the 2nd and 3rd term, 3rd and 4th term, and so on, we can see that each of these terms are negative.
∴ The sum is less than 1.
The only option that satisfies this is .
Hence, option (d).
Workspace:
If x2 + 3x – 10 is a factor of 3x4 + 2x3 – ax2 + bx – a + b – 4, then the closest approximate values of a and b are
- (a)
25, 43
- (b)
52, 43
- (c)
52, 67
- (d)
None of the above
Answer: Option C
Text Explanation :
x2 + 3x – 10 has 2 factors 2 and −5
x2 + 3x – 10 is a factor of 3x4 + 2x3 – ax2 + bx – a + b – 4
Therefore 2 and −5 are also the factors of this expression.
Substitute 2 and −5, to get 2 equations
For x = 2,
48 + 16 – 4a + 2b – a + b – 4 = 0
∴ 5a – 3b = 60…… (I)
For x = −5,
1875 – 250 – 25a – 5b – a + b – 4 = 0
26a + 4b = 1621……… (II)
Solving (I) and (II)
a = 52 and b = 67
Hence, option (c).
Workspace:
If x is real, the smallest value of the expression 3x2 – 4x + 7 is:
- (a)
2/3
- (b)
3/4
- (c)
7/9
- (d)
None of the above
Answer: Option D
Text Explanation :
The minimum value of expression ax2 + bx + c
= = =
Hence, option (d).
Workspace:
Nikhil’s mother asks him to buy 100 pieces of sweets worth Rs. 100/-. The sweet shop has 3 kinds of sweets, kajubarfi, gulabjamun and sandesh. Kajubarfi costs Rs. 10/- per piece, gulabjamun costs Rs. 3/- per piece and sandesh costs 50 paise per piece. If Nikhil decides to buy at least one sweet of each type, how many gulabjamuns should he buy?
- (a)
1
- (b)
2
- (c)
3
- (d)
4
- (e)
5
Answer: Option A
Text Explanation :
Let Nikhil buy a, b and c pieces of kajubarfi, gulabjamun and sandesh respectively.
Hence, we have,
a + b + c = 100, and … (i)
10a + 3b + 0.5c = 100 … (ii)
By, 2 × ii – i, we get,
19a + 5b = 100
Hence, a = (100 – 5b)/19
Now, (100 – 5b)/19 will be positive integer only if b = 1.
In that case, a = 5.
Hence, Nikhil must buy 1 gulabjamun.
Hence, option (a).
Workspace:
A potter asked his two sons to sell some pots in the market. The amount received for each pot was same as the number of pots sold. The two brothers spent the entire amount on some packets of potato chips and one packet of banana chips. One brother had the packet of banana chips along with some packets of potato chips, while the other brother just had potato chips. Each packet of potato chips costs Rs. 10/- and the packet of banana chips costs less than Rs. 10/-. The packets of chips were divided between the two brothers so each brother received equal number of packets. How much money should one brother give to the other to make the division financially equitable?
- (a)
1
- (b)
2
- (c)
4
- (d)
5
- (e)
7
Answer: Option B
Text Explanation :
Let the potter’s sons have x pots.
Hence, they received Rs. x2 after selling these pots.
As the price of one banana wafer packet is less than Rs. 10 hence, x2 will not be a multiple of 100.
Assume that they bought n packets of potato wafers.
Hence, total number of wafers packet = n + 1
Hence, each son gets (n + 1)/2 packets.
Hence, n is odd.
Let b be the price of banana wafers.
Hence, they have Rs. (n × 10 + b)
As n is odd, tens place of (n × 10 + b) is odd.
Now, each brother can have equal money if total amount earned by them is even.
Hence, b must be even.
Hence, we have the following condition,
x2 = 10 × n + b such that b is even and n is odd.
Hence, x is an even number.
Now, if unit’s digit is of x is 2 or 8, then tens place of x2 will be even.
This is violates our condition that tens digit of x2 is odd and hence it is not possible.
Hence, unit’s place digit of x is 4 or 6.
In either case, unit digit of x2 is 6.
Hence, b = 6.
Hence, the son having banana wafers owes Rs. ((n – 1) × 10/2 + 6) and the other son owes ((n + 1) × 10/2)
Hence, one of the son has Rs. ((n + 1) × 10/2) – ((n – 1) × 10/2 + 6) = Rs. 4 more than the other.
Hence, he must give Rs. 2 to the other to have financially equitable division.
Hence, option (b).
Workspace:
Three truck drivers, Amar, Akbar and Anthony stop at a road side eating joint. Amar orders 10 rotis, 4 plates of tadka, and a cup of tea. Akbar orders 7 rotis, 3 plates of tadka, and a cup of tea. Amar pays Rs. 80 for the meal and Akbar pays Rs. 60. Meanwhile, Anthony orders 5 rotis, 5 plates of tadka and 5 cups of tea. How much (in Rs.) will Anthony pay?
- (a)
75
- (b)
80
- (c)
95
- (d)
100
- (e)
None of the above
Answer: Option D
Text Explanation :
Let cost of a Roti, a plate of Tadka and a cup of tea be Rs. a, b and c respectively.
Hence, we have,
10a + 4b + c = 80, and … (I)
7a + 3b + c = 60 … (II)
Now, by II – I,
3a + b = 20 … (III)
Hence, by, (I) – 3(III), we have,
a + b + c = 20
Hence, 5a + 5b + 5c = 100
Hence, Anthony will pay Rs. 100.
Hence, option (d).
Workspace:
A computer program was tested 300 times before its release. The testing was done in three stages of 100 tests each. The software failed 15 times in Stage I, 12 times in Stage II, 8 times in Stage III, 6 times in both Stage I and Stage II, 7 times in both Stage II and Stage III, 4 times in both Stage I and Stage III, and 4 times in all the three stages. How many times the software failed in a single stage only?
- (a)
10
- (b)
13
- (c)
15
- (d)
17
- (e)
35
Answer: Option B
Text Explanation :
Assume that the software fails a, b and c times in a single stage, in two stages and in all stages respectively.
Hence, b + 3c = 6 + 7 + 4 = 17
But c = 4, hence, b = 5
Similarly, we have,
a + 2b + 3c = 15 + 12 + 8
Hence, a = 13
Hence, option (b).
Workspace:
The equation 7x – 1 + 11x – 1 = 170 has
- (a)
no solution
- (b)
one solution
- (c)
two solutions
- (d)
three solutions
Answer: Option B
Text Explanation :
7x – 1 + 11x – 1 = 170
WE can see that the RHS is a multiple of 10.
11x – 1 has 1 in its units place.
∴ 7x – 1 should have 9 in its units place.
The lowest value for which this is true is x = 3
72 + 112 = 170
We can see that the for any other value of x, which is greater than 3,
7x – 1 + 11x – 1 > 170
Hence, option (b).
Workspace:
The value of is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Answer: Option C
Text Explanation :
Let x =
∴ = x
∴ 7 + = x2
Substituting the options, only x = 3 satisfies the equation.
Hence, option (c).
Workspace:
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