MAT Mock Test - 10 - MAT MCQ

She goes for a walk every morning to the park.

As per the rule of subject-verb agreement, singular verb is used with singular subject.

The use of the singular verb 'goes' is appropriate in the space of the sentence. 

The idiom "sow wild oats" is used for young people who waste their time doing stupid or idle things OR to have many sexual relationships particularly when one is young.

Example: They look upon his indiscretions as just his sowing his wild oats.

Let's look at the meaning of the given idiom:

A Hard nut to crack - a problem that is very difficult to solve or a person who is very difficult to understand. For Example: The test problem was a hard nut to crack.

The first statement is B as mentioned and it introduces us to the topic of the passage which is a particular personality. F follows B as it talks about her appointment. C follows F as it mentions a reaction of a person to her appointment. E is the fourth statement as mentioned in the passage. D is the fifth statement as it talks on the same note as that of E. A is the sixth statement and G follows A as the words ‘in return’ are mentioned here.
Thus the correct order is BFCEDAG.
The FIFTH statement is D.

The word 'harassed' refers to someone that is strained by an external factor troubling them.
The word 'assisted' refers to someone that is helped along by someone else.
Thus, it is clear that 'harassed' and 'assisted' are opposite words.

As per the English alphabetical series and its place value.

The logic followed here is:

D + 2 = F; F + 2 = H; H + 2 = J; J + 2 = L

A + 2 = C; C + 2 = E; E + 2 = G; G + 2 = I

G + 2 = I; I + 2 = K; K + 2 = M; M + 2 = O

So, “HEK” is the correct answer.

The pattern followed is,

Thus, OP is the correct answer.

Box colours = Red, Green, White, Pink, Brown, Orange, Blue, and Yellow
1) Four boxes should be kept between the Brown and the Orange box.
2) Either the Orange or the Brown box should be placed at the bottom.

3) The Red box should be placed three boxes above the Brown box.
(Implies, the Brown box is at the bottom only then we can place the Red box three boxes above the Brown box.)

4) At least five boxes should be placed below the Green box.
5) The Blue box can’t be placed below the Green box.
(As the Blue box is kept above the Green box. Implies, the Blue box is kept at the top and the Green box is kept immediately below the Blue box.)

6) The Pink box should not be placed next to the Red box.
(There is only one way to place the Pink box, i.e. immediately above the Brown box.)

7) The White box cannot be placed above the Yellow box.
(As the White box is placed below the Yellow box, implies, the White box is kept immediately above the Pink box and the Yellow box is kept immediately above the Red box.)

Clearly, five boxes are kept above the White box.

Box colours = Red, Green, White, Pink, Brown, Orange, Blue, and Yellow
1) Four boxes should be kept between the Brown and the Orange box.
2) Either the Orange or the Brown box should be placed at the bottom.

3) The Red box should be placed three boxes above the Brown box.
(Implies, the Brown box is at the bottom only then we can place the Red box three boxes above the Brown box.)

4) At least five boxes should be placed below the Green box.
5) The Blue box can’t be placed below the Green box.
(As the Blue box is kept above the Green box. Implies, the Blue box is kept at the top and the Green box is kept immediately below the Blue box.)

6) The Pink box should not be placed next to the Red box.
(There is only one way to place the Pink box, i.e. immediately above the Brown box.)

7) The White box cannot be placed above the Yellow box.
(As the White box is placed below the Yellow box, implies, the White box is kept immediately above the Pink box and the Yellow box is kept immediately above the Red box.)

Clearly, the Pink box is kept immediately above the Brown box.

It follows the pattern as:

L + 4 = P, P + 4 = T, T + 4 = X

L – 3 = I, I – 3 = F, F – 3 = C

N + 2 = P, P + 2 = R, R + 2 = T

O – 1 = N, N - 1 = M, M – 1 = L

Thus, PIPN completes the series.

Given,
Total number of students = 27
The rank of X from right end =14
Rank of X from left = Total number of students - Rank of X from right +1
= (27 - 14 + 1)
= 14

Length of Train P = 175 m

Let the speed of the train be y.

Length of Train Q = 150 m

Let the speed of the train be x.

For Statement I:

325 = (x + y) × 13

⇒ x + y = 25      ----(i)

Statement I alone is not sufficient to answer the question.

For statement II: 

325 = (x - y) × 65

⇒ x - y = 5      ----(ii)

Statement II alone is not sufficient to answer the question.

But if we combine both statements I and II then we can find the Speed.

For statement III:

x + y = 25      ----(iii)

Statement III alone is not sufficient to answer the question.

But if we combine statements II and III we can find the Speed.

∴ Either statement I and II together or Statement II and III together is sufficient to answer the question.

From statement I:
The selling price of the milk = 2 × Cost price of milk Profit

From statement II:
Let the C.P. of pure milk per liter be x.
8 liters of mixture = 2 liters of water +6 liters of milk
Cost price of 8 liters of mixture = C.P. of 2 liter water + C.P. of 6 liter milk = 0 + 6x = 6x
From I, S.P. of 8 liter of mixture = S.P. of 8 liters of pure milk  = 8 × 2x = 16x

Profit%
∴The data in either statement I alone or in statement II alone are sufficient to answer the question.

From statement I,

Let, speed of Sandeep and Prateek be ‘x’ km/hr and ‘y’ km/hr respectively.

Let, time taken by both to meet each other be ‘T’ hours.

Distance between point A to point C = 9y km

Distance between point B to point C = 4x km

So, T = 9y/x

Also, T = 4x/y

⇒ T × T = (9y/x) x (4x/y)

⇒ T² = 36

⇒ T = 6 hours

Then, both will meet each other at 9:00 pm.

Thus, Statement I is sufficient to answer.

From statement II,

Let, speed of Sandeep and Prateek be 2a km/hr and 3a km/hr respectively.

⇒ Time taken to meet each other = 1200/5a

So, statement II is not sufficient

From statement III,

Difference between the speed of Sandeep and Prateek is 40 km/hr.

Nothing can be concluded from this statement.

By combining II and III,

3a - 2a = 40

⇒ a = 40 km/hr

Speed of Sandeep and Prateek will be 80 km/hr and 120 km/hr respectively.

⇒ Time taken to meet each other = 1200/200 = 6 hours

So, both will meet each other at 9:00 pm.

∴ Either I only or II and III together are sufficient to answer.

Total quantity of Food items used in the month of February =  250 + 380 + 350 + 310 = 1290
Total quantity of Food items used in the month of March =  230 + 370 + 290 + 300 = 1190
Required Ratio = 1290 : 1190 =  129 : 119

Number of students enrolled in 2019 in both camps = 70 + 150 = 220
Total number of students enrolled in 2020 in both camps = 130x220/100 = 286

Given:
Percentage of passengers travelling from Haridwar to Delhi, Rishikesh, Jaipur and Amritsar respectively = 20%, 19%, 10%, and 11%
Total numbers of passengers = 1500
Total percentage of passengers travelling from Haridwar to Delhi, Rishikesh, Jaipur and Amritsar = 20% + 19% + 10% + 11%
= 60%
So the number of passengers travelling from Haridwar to Delhi, Rishikesh, Jaipur and Amritsar = 60% of 1500
= 1500 x 60/100
= 900
Average = 
= 900/4
= 225
∴ The average number of passengers travelling between Haridwar to Delhi, Rishikesh, Jaipur and Amritsar is 225.

Given:
Percentage of passengers travelling from Haridwar to Moradabad = 25%

Central angle for pie chart =  x 360º
Central angle for passengers = 25% × 360°
= 25/100 x 360º
= 90º
∴ Central angle for passengers travelling from Haridwar to Moradabad = 90°

Given:

25 cosec² x + 36 sec² x

As we know that, minimum value of 

Here, m = 36 and n = 25.

∴ The minimum value of 25 cosec² x + 36 sec² x : 

Let AF = FB = a and BD = DC = b

In ΔABD, AD² = AB² + BD² ... (1)

⇒ AD² = 4a² + b²

In ΔBFC, FC² = FB² + BC²

⇒ FC² = a² +4 b² ...(2)

Adding equation 1 and 2 ,

⇒ AD² + FC² = 5a² +5 b²

Also, AC² = 4a² + 4b² 

⇒ 4 (AD² + FC²) = 5AC²

Let the number of spectators on Monday, Tuesday, and Wednesday be 2x, 3x, and 5x respectively.

Let the price charged on Monday, Tuesday, and Wednesday be 2y, 3y, and 4y respectively.

According to the question

⇒ (5X × 4Y) - (3X × 3Y) = 8800

⇒ XY = 800

Now, the total amount earned on all three days

⇒ (2X × 2Y) + (3X × 3Y) + (5X × 4Y) 

⇒ XY × (4 + 9 + 20) = 33XY = 33 × 800 = 26400

∴ The required result will be 26400.

Given:
A, B and C together can complete a work = 15 days
Efficiency of A is 20% more than B.
Efficiency of B is 25% more than C.
Total work = Efficiency × Time
Let efficiency of C be 4 units/day
Efficiency of B = 4 × 
= 5 units/day

Efficiency of A = 5 × 
= 6 units/day
Total work = (4 + 5 + 6) × 15
= 15 × 15 
= 225
∴ Time taken by B to complete whole work = 225/5
= 45

Let second-degree equation be

It will represent a pair of straight lines, If discriminant (Δ) of this equation equal to zero (Δ = 0) 

Discriminant 

Given: Second degree equation,

Compare with second-degree equation

So, and

Given equation represents a pair of straight lines

So,

Rate of flow of water in each pipe =  = 36 : 16 : 81

Part of the cistern filled by pipe A in 1 min = 36k

Part of the cistern filled by pipe B in 1 min = 16k

And that filled by C in 1 min = 81k

The largest pipe fills the tank in 133 min.

So, cistern filled by C in 133 min = 133 × 81k = 1

k 

Now part of the cistern filled by all the three in 1 min = 36k + 16k + 81k = 133k 

So, all three can fill the cistern in 81 min.

Combination is the number of ways in which we can select a group of objects from a set.

For example, selecting r things from a set of n things is given as .

We have to select 11 players from a total of 15 players.

So, the number of ways

The landscape is being continuously worn away by two processes – weathering and erosion. Erosion is the wearing away of the landscape by different agents like water, wind, and ice.

• At higher gradients, downward, vertical erosion is more dominant. This produces V-shaped valleys
• The V-shaped valley is typical of one that has been carved by flowing water.
• The erosion is more pronounced when the water flow is a heavy one, and the water carries suspended particles (sedimentary load).

Agricultural Produce Market Committee (APMC) is a statutory market committee constituted by a State Government. 
It is constituted for trade in certain notified agricultural or horticultural or livestock products, under the Agricultural Produce Market Committee Act issued by that state government. 

The fifth edition of the Khelo India Youth Games has started in Bhopal in the presence of Madhya Pradesh Chief Minister Shivraj Singh Chouhan.

Bihar has enforced a mandatory requirement for investigators to conclude case investigations within 75 days of FIR registration, effective from January 1 next year, as part of enhanced policing measures.

Chlamydomonas is an example of a green algae.

Chlamydomonas is found in stagnant water and on damp soil as well as freshwater, seawater, and even in snow. It is found almost everywhere and consists of 325 species.